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Principles of Organic Synthesis, Third Edition (Norman, Richard O.C.; Coxon, James M.)

2018-01-13T22:00:00.000-08:00

Principles of Organic Synthesis, Third Edition (Norman, Richard O.C.; Coxon, James M.)

Shelton Bank

State University of New York at Albany, Albany, NY 12222

J. Chem. Educ., 1995, 72 (7), p A152

DOI: 10.1021/ed072pA152.1

Publication Date: July 1995

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Organic Chemistry, 2 edition by Jonathan Clayden,

2018-01-13T21:41:00.000-08:00

Organic Chemistry, 2 edition by Jonathan Clayden, Nick Greeves and Stuart Warren

Topics reaction, chapter, reactions, bond, acid, carbonyl, group, compounds, carbon, bonds, carbonyl group, double bond, leaving group, conjugate addition, lone pair, carbon atom, nmr spectrum, starting material, interactive mechanism, carbonyl compounds

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By SAJJAD HUSSAIN MIRANI
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Converting between "ppm" and molarity

2016-05-27T02:36:00.001-07:00

Converting between "ppm" and molarity

The definition of parts per million:

1 g solute per 1,000,000 g solution
Now, divide both values by 1000 to get a new definition for ppm:

ppm = 0.001 g per 1,000 g solution
or:

ppm = 1 mg solute per 1 kg solution
Then, for an aqueous solution:

ppm = 1 mg solute per liter of solution
This last one works because the solution concentration is so low that we can assume the solution density to be 1.00 g/mL.

Also, it's this last modification of ppm (the mg/L one) that allows us to go to molarity (which has units of mol/L).

The best way to explain this is by doing some examples:

Problem #1: Convert 78 ppm of Ca2+ ions to mol/L

Solution:

1) By the last definition of ppm just above:

78 ppm = 78 mg Ca2+ / L of solution = 0.078 g/L
2) Divide by the atomic weight for calcium ion:

0.078 g/L divided by 40.08 g/mol = 00019 mol/L
Problem #2: Calculate the molarity of a dye concentration given the molar mass is of the dye 327 g/mol and a dye concentration of 2 ppm.

Solution:

1) Convert ppm to a gram-based concentration:

2 ppm = 2 mg dye / L of solution
2a) Using 0.002 g/L, caculate the molarity:

0.002 g/L divided by 327 g/mol = 6.1 x 10-6 M
2b) Using 2 mg/L, calculate the molarity

2 mg/L divided by 327,000 mg/mol = 6.1 x 10-6 M
You might want to go back to problem #1 and try out 78 mg/L with the atomic weight of calcium ion expressed as mg/mol instead of g/mol.

Problem #3: A solution is labeled 2.89 ppm and is made with a solute that has molar mass equal to 522 g/mol. What is the molarity of the solution?

Solution (straight from Yahoo Answers):

1) Unless specified otherwise, ppm usually refers to ppm by weight:

2.89 ppm = 2.89 g per 1,000,000 g (or any other weight unit, I have just chosen g for convenience)
2) Now you need to know the density of the solvent to convert the volume to mass. If the solvent is water, we can assume the density at standard temperature and pressure is 1.0 g/mL. Therefore 1 litre (L) of water is 1,000 g. Therefore:

2.89 ppm = 2.89 g per 1,000 L = 0.00289 g per 1 L
(If you work out the mass per litre it makes working out the next steps a little easier, because the final units for molarity will be mol/L)

3) Now using the molar mass to work out the molarity (moles per litre):

Moles in one litre = mass/molar mass = 0.00289/522 = 5.5 x 10-6 mol
Therefore the molarity is 5.5 x 10-6 M

Problem #4: What is the ppm concentration of calcium ion in 0.010 M CaCO3?

Solution:

1) Convert mol/L to to g/1000 g of solution:

0.010 mol/L times 40.08 g/mol = 0.4008 g/L
0.4008 g/L = 0.4008 g / 1000 g of solution

Note that the solution density is assumed to be 1.00 g/mL. This is the common practice in ppm calculations.

2) Convert to ppm by multiplying numerator and denominator by 1000:

0.4008 g / 1000 g of solution times 1000/1000 = 400 g / 1,000,000 g of solution
The answer is 400 ppm

Notice how the weight for calcium ion only is used. The weight of the carbonate, based on the wording of the question, is immaterial to the solving of the problem.

This focus on an ion (and not the entire formula of the substance) in ppm concentrations is common. Be aware!

Problem #5: Make a 20.0 ppm solution of HCl from a stock solution that is 0.500-molar.

Solution:

1) Let us assume we will make 1.00 L of solution. Since ppm = 1 mg solute per liter of solution:

we need 20.0 mg solute per liter of solution
2) Calculate the volume of 0.500 M solution that contains 20.0 mg HCl:

MV = g/molar mass
(0.500 mol/L) (x) = 0.020 g / 36.46 g/mol

x = 0.001097 L

We need 1.10 mL of the stock solution diluted to 1.00 L to make a 20.0 ppm solution of HCl.

3) A slightly different form of the above question was answered thusly on Yahoo Answers:

The unit "ppm" is equivalent to "mg/L."
Concentrated HCl is 12.3 moles/L or, since the molar mass of HCl is 36.5 g/L, 449 g HCl / L.

449 g/L = 449,000 mg/L. If you want 20 ppm, you need to make a 20/449,000 volume dilution, or 1/22,500.

The best way to do that is a serial dilution: first make a 1/100 dilution (1 mL conc. HCl / 100 mL) and then dilute THAT solution by 1/225. The result: 1/100 x 1/225 = 1/22,500.

You can cut the dilution process to one step if a lower concentration of HCl is available (e.g. 0.1 M).

Problem #6: In a geographical area at sea level measuring 56 km by 24 km by 1 km, the ozone concentration is 0.145 ppm. How many moles of ozone must be removed from the atmosphere to reach a concentration of 0.080 ppm, assuming a constant pressure of 1 atm and a constant temperature of 30 degrees Celsius.

Solution:

1) Calculate the liters (use cubic decimeters) in our volume:

(56 x 105 dm) (24 x 105 dm) (1 x 105 dm) = 1.344 x 1018 dm3
2) Calculate the ppm that must be removed:

0.145 - 0.080 = 0.065
3) Determine milligrams (then grams, then moles) of ozone to be removed from our volume:

1 ppm = 1 mg/L
Therefore:

0.065 mg/L times 1.344 x 1018 L = 8.736 x 1016 mg

8.736 x 1016 mg = 8.736 x 1013 g

8.736 x 1013 g divided by 47.997 g/mol = 1.75 x 1012 mol

Problem #7: 3100 mL of a river water sample took 9.30 mL of 0.01005 M Ag+ to titrate. Calculate the concentration of Cl¯ in ppm for the river water.

Solution:

1) Calculate moles of Ag+:

(0.01005 mol/L) (0.00930 L) = 9.3465 x 10-5 mol of Ag+
2) Since Ag+ and Cl¯ react in a 1:1 molar ratio:

9.3465 x 10-5 mol of Cl¯
3) How many mg of Cl¯ is this?

9.3465 x 10-5 mol x 35.453 g/mol = 3.31 x 10-3 g
3.31 x 10-3 g x (1000 mg/g) = 3.31 mg

4) ppm = mg/L

3.31 mg / 3.100 L = 1.07 ppm (to three sf)
Problem #8: A solution contains 6.0 x 10-6 mol Na2SO4 in 250.0 mL. What is the concentration of sodium ion in ppm? Of sulfate ion?

Solution:

1) Convert to molarity:

6.0 x 10-6 mol / 0.250 L = 2.40 x 10-5 mol/L
2) Convert mol/L to grams of sodium ion per 1000 g of solution:

2.40 x 10-5 mol/L times 23.00 g/mol x 2 = 0.001104 g/L
0.001104 g/L = 0.001104 g/1000 g of solution

The two is used because there are two sodium ions per formula unit of sodium sulfate.

For sulfate, the two would not be used and, instead of 23.00 g/mol, use the "formula weight" of the sulfate ion.

3) Convert to ppm by multiplying numerator and denominator by 1000:

0.001104 g/1000 g of solution times 1000/1000 = 1.104 g / 1,000,000 g of solution
The answer (for sodium ion) is 1.1 ppm (to two sig figs).

The solution for sulfate is left to the reader.

Problem #8: The maximum concentration of O2 in seawater is 2.2 x 10-4 M at 25.0 °C. What is this concentration in ppm?

Solution:

1) Convert mol/L to grams of sodium ion per 1000 g of solution:

2.2 x 10-4 mol/L times 32.0 g/mol = 0.00704 g/L
0.00704 g/L = 0.00704 g/1000 g of solution

2) Convert to ppm by multiplying numerator and denominator by 1000:

0.00704 g/1000 g of solution times 1000/1000 = 7.04 g / 1,000,000 g of solution
The answer is 7.0 ppm (to two sig figs).

What does "ppm" mean?

2016-05-27T02:35:00.000-07:00

What does "ppm" mean?
Go To Converting between "ppm" and molarity

The expression "1 ppm" means a given solute exists at a concentration of one part per million parts of the solution. These are two common ways to think about what the concentration "1 ppm" means:

a) it is one-millionth of a gram per gram of sample solution.
b) it is one gram of solute per million grams of sample solution.

Notice that the more general word 'part' is used above, but 'gram' is used in (a) and (b) just above. This is because 'gram' is used almost exclusively when parts per million is used.

The best way to explain this is by doing some examples:

Problem #1: Sea water contains 3.90 x 10¯6 ppm of dissolved gold. What volume of this sea water would contain 1.00 g of gold?

Solution:

1) 3.90 x 10¯6 ppm means this:

3.90 x 10¯6 g of Au per 1.00 gram of seawater
2) We use a ratio and proportion:

(3.90 x 10¯6 g of Au ÷ 1.00 gram of seawater) = (1.00 g of Au ÷ x)
3) Cross multiply and divide yields our answer:

2.56 x 105 g of sea water contains 1.00 g of gold
Problem #2: Pollutants in air and water are frequently measured in parts per million (ppm) or parts per billion (ppb). One part per million would mean that there is one gram of the pollutant in every one million grams of air. At ordinary temperature and pressure, air has a density of 0.00012 gram per cubic centimeter. What volume of air would contain one gram of sulfur dioxide, a pollutant that causes acid rain, if the sulfur dioxide concentration is 2 ppm.

Solution:

1) How many grams of air contain one gram of SO2?

2 ppm means 2 grams SO2 per million grams of air
therefore,

one gram of SO2 will be found in 500,000 g of air

2) What volume of air weighs 500,000 g? We use the density (notice I set it up in a ratio and proportion style:

0.00012 g over 1 cm3 = 500,000 g over x
x = 4.17 x 109 cm3

Problem #3: A sample of oil (density = 0.89 g/mL) was found to have dioxin contamination of 2 ppm. How many mL of the oil would contain 0.01 gram of dioxin?

Solution:

1) Determine grams of oil holding 0.01 g of dioxin. Use a ratio and proportion:

(2 g dioxin over 1,000,000 g solution) = (0.01 g over x)
x = 5000 g

2) Determine the volume of oil:

0.89 g/mL = 5000 g divided by x
x = 5618 mL

Problem #4: Fish need about 5 ppm O2 dissolved in water to survive. Will water with 7 mg O2 per liter sustain fish?

Solution:

We need to convert 7 mg/L into ppm:

(0.007 g / 1000 g) = (x / 1,000,00 g)
x = 7 ppm

The answer is yes.

Notice the conversion of one liter of water (with 7 mg dissolved oxygen) into 1000 g. I used the density of 1.00 g/mL for the conversion. This is acceptable because solutions in the ppm range are so dilute that the solute (in this case, the O2, has no effect on the solution density.

Look again at the mg/L value. Another way to explain getting from mg/L to ppm is to multiply both numerator and denominator by 1000:

(0.007 g / 1000 g) x (1000 / 1000) = 7 g / 1,000,000 g = 7 ppm
You may very well face problems where you get the data in mg/L format and are then asked something about ppm. Remember!

Problem #5: Show why a mass of 0.145 g of KMnO4 in a 500 mL volumetric flask corresponds to 100 ppm in Mn.

Solution:

1) Using a gravimetric factor, determine grams of Mn in 0.145 g of KMnO4:

0.145 g times (54.938 g/mol over 158.032 g/mol) = 0.0504076 g
2) Calculate ppm of Mn. Use a ratio and proportion:

(0.050 g over 500 g) = (x over 1,000,000)
x = 100 ppm

Problem #6: Symptoms of lead poisoning become apparent after a person has accumulated more than 20 mg in the body. Express this amount as parts per million for an 80 kg person.

Solution:

1) Convert mg and kg to grams:

20 mg = 20 x 10-3 g
80 kg = 80 x 103 g
2) Determine grams Pb per g of bodyweight:

20 x 10-3 g / 80 x 103 g = 0.25 x 10-6 per g of bodyweight
3) Refer to first description of ppm given at beginning of file:

one ppm "is one-millionth of a gram per gram of sample solution."
In this case, the "solution" is the bodyweight.

answer = 0.25 ppm

Problem #7: If you eat a 6 oz. can (180g) of tuna with 0.20 ppm Hg, how much mercury do you ingest?

Solution:

1) The meaning of 0.20 ppm Hg:

Remember, one ppm "is one-millionth of a gram per gram of sample solution."
Therefore the tuna contains 0.20 x 10-6 g of Hg per gram of tuna.

2) Calculate Hg in 180 g of tuna:

0.20 x 10-6 g of Hg per g of tuna times 180 g of tuna = 3.6 x 10-5 g of Hg
Problem #8: It is estimated that 3 x 105 tons of sulfur dioxide enters the atmosphere daily owing to the burning of coal and petroleum products. Assuming an even distribution of the sulfur dioxide throughout the earth's atmosphere (which is not the case), calculate in parts per million by weight the concentration of SO2 added daily to the atmosphere. The weight of the atmosphere is 4.5 x 1015 tons.

Solution:

1) Set up a ratio and proportion:

You are adding 3 x 105 parts per 4.5 x 1015 parts.
How many parts per 1 x 106 is this?

2) Solve it:

3 x 105 parts is to 4.5 x 1015 parts as x is to 1 x 106 parts
x = 0.000067 ppm

Problem #9: A solution used to chlorinate a home swimming pool contains 7% chlorine by mass. An ideal chlorine level for the pool is one part per million chlorine. If you assume densities of 1.10 g/mL for the chlorine solution and 1.00 g/mL for the swimming pool water, what volume of the chlorine solution in liters, is required to produce a chlorine level of 1.00 ppm in an 18,000 gallon swimming pool?

Solution:

1) Convert 18,000 gallons to liters:

The conversion factor we will use is 1 gallon = 3.7854 L
18,000 gal x 3.7854 L/gal = 6.81372 x 104 L

2) Determine how many grams of pool water this is:

6.81372 x 107 mL x 1.00 g/mL = 6.81372 x 107 g
Note change from L to mL.

3) At 1 ppm, how much chlorine is required? Use a ratio and proportion:

(1 g chlorine ÷ 106 g pool water) = (x ÷ 6.81372 x 107 g of pool water)
x = 68.1372 g chlorine required

4) What amount of 7% (by mass) chlorine solution is required to deliver 68.1372 g of chlorine?

(68.1372 g ÷ 0.07) = (x/1)
x = 973.3886 g of chlorine solution required

5) What volume (in liters) is this?

973.3886 g ÷ 1.10 g/mL = 884.8987 mL
To three sig figs (which seems reasonable to the ChemTeam), the answer is 0.885 L.

Molarity Problems

2016-05-27T02:31:00.000-07:00

Molarity Problems
Molarity Difination
The equations I will use are:

M = moles of solute / liters of solution
and

MV = grams / molar mass <--- The volume here MUST be in liters.
Typically, the solution is for the molarity (M). However, sometimes it is not, so be aware of that. A teacher might teach problems where the molarity is calculated but ask for the volume on a test question.

Note: Make sure you pay close attention to multiply and divide. For example, look at answer #8. Note that the 58.443 is in the denominator on the right side and you generate the final answer by doing 0.200 times 0.100 times 58.443.

Problem #1: Sea water contains roughly 28.0 g of NaCl per liter. What is the molarity of sodium chloride in sea water?

Solution:

MV = grams / molar mass
(x) (1.00 L) = 28.0 g / 58.443 g mol¯1

x = 0.4790993 M

to three significant figures, 0.479 M

Problem #2: What is the molarity of 245.0 g of H2SO4 dissolved in 1.000 L of solution?

Solution:

MV = grams / molar mass
(x) (1.000 L) = 245.0 g / 98.0768 g mol¯1

x = 2.49804235 M

to four sig figs, 2.498 M

If the volume had been specified as 1.00 L (as it often is in problems like this), the answer would have been 2.50 M, NOT 2.5 M. You want three sig figs in the answer and 2.5 is only two SF.

Problem #3: What is the molarity of 5.30 g of Na2CO3 dissolved in 400.0 mL solution?

Solution:

MV = grams / molar mass
(x) (0.4000 L) = 5.30 g / 105.988 g mol¯1

0.12501415 M

x = 0.125 M (to three sig figs)

Problem #4: What is the molarity of 5.00 g of NaOH in 750.0 mL of solution?

Solution:

MV = grams / molar mass
(x) (0.7500 L) = 5.00 g / 39.9969 g mol¯1

(x) (0.7500 L) = 0.1250097 mol <--- threw in an extra step

x = 0.1666796 M

x = 0.167 M (to three SF)

Problem #5: How many moles of Na2CO3 are there in 10.0 L of 2.00 M solution?

Solution:

M = moles of solute / liters of solution
2.00 M = x / 10.0 L

x = 20.0 mol

Suppose the molarity was listed as 2.0 M (two sig figs). How to display the answer? Like this:

20. mol

Problem #6: How many moles of Na2CO3 are in 10.0 mL of a 2.0 M solution?

Solution:

M = moles of solute / liters of solution
2.0 M = x / 0.0100 L <--- note the conversion of mL to L

x = 0.020 mol

Problem #7: How many moles of NaCl are contained in 100.0 mL of a 0.200 M solution?

Solution:

0.200 M = x / 0.1000 L
x = 0.0200 mol

Problem #8: What weight (in grams) of NaCl would be contained in problem #7?

Solution:

(0.200 mol L¯1) (0.100 L) = x / 58.443 g mol¯1 <--- this is the full set up
x = 1.17 g (to three SF)

You could have done this as well:

58.443 g/mol times 0.0200 mol <--- this is based on knowing the answer from problem #7

Problem #9: What weight (in grams) of H2SO4 would be needed to make 750.0 mL of 2.00 M solution?

Solution:

(2.00 mol L¯1) (0.7500 L) = x / 98.0768 g mol¯1
x = (2.00 mol L¯1) (0.7500 L) (98.0768 g mol¯1)

x = 147.1152 g

to three sig figs, 147 g

Problem #10: What volume (in mL) of 18.0 M H2SO4 is needed to contain 2.45 g H2SO4?

Solution:

(18.0 mol L¯1) (x) = 2.45 g / 98.0768 g mol¯1
(18.0 mol L¯1) (x) = 0.0249804235 mol

x = 0.0013878 L

The above is the answer in liters. Multiplying the answer by 1000 provides the required mL value:

0.0013878 L times (1000 mL / L) = 1.39 mL (given to three sig figs)

Problem #11: What volume (in mL) of 12.0 M HCl is needed to contain 3.00 moles of HCl?
Solution:

12.0 M = 3.00 mol / xx = 0.250 L
This calculates the volume in liters. Multiplying the answer by 1000 provides the required mL value:
0.250 L x (1000 mL / L) = 250. mL (note use of explicit decimal point to create three sig figs)

Problem #12: How many grams of Ca(OH)₂ are needed to make 100.0 mL of 0.250 M solution?
Solution:

(0.250 mol L¯¹) (0.100 L) = x / 74.0918 g mol¯¹x = (0.250 mol L¯¹) (0.100 L) (74.0918 g mol¯¹)
x = 1.85 g (to three sig figs)

Problem #13: What is the molarity of a solution made by dissolving 20.0 g of H₃PO₄ in 50.0 mL of solution?
Solution:

(x) (0.0500 L) = 20.0 g / 97.9937 g mol¯¹(x) (0.0500 L) = 0.204094753 mol
x = 4.08 M

Problem #14: What weight (in grams) of KCl is there in 2.50 liters of 0.500 M KCl solution?
Solution:

(0.500 mol L¯¹) (2.50 L) = x / 74.551 g mol¯¹x = 93.2 g

Problem #15: What is the molarity of a solution containing 12.0 g of NaOH in 250.0 mL of solution?
Solution:

(x) (0.2500 L) = 12.0 g / 39.9969 g mol¯¹ <--- note use of L in set up, not mL x = 1.20 M

Problem #16: Determine the molarity of these solutions:

a) 4.67 moles of Li₂SO₃ dissolved to make 2.04 liters of solution.
b) 0.629 moles of Al₂O₃ to make 1.500 liters of solution.
c) 4.783 grams of Na₂CO₃ to make 10.00 liters of solution.
d) 0.897 grams of (NH₄)₂CO₃ to make 250 mL of solution.
e) 0.0348 grams of PbCl₂ to form 45.0 mL of solution.

Solution set-ups:

a) x = 4.67 mol / 2.04 L
b) x = 0.629 mol / 1.500 L
c) (x) (10.00 L) = 4.783 g / 106.0 g mol¯¹
d) (x) (0.250 L) = 0.897 g / 96.09 g mol¯¹
e) (x) (0.0450 L) = 0.0348 g / 278.1 g mol¯¹

Problem #17: Determine the number of moles of solute to prepare these solutions:

a) 2.35 liters of a 2.00 M Cu(NO₃)₂ solution.
b) 16.00 mL of a 0.415-molar Pb(NO₃)₂ solution.
c) 3.00 L of a 0.500 M MgCO₃ solution.
d) 6.20 L of a 3.76-molar Na₂O solution.

Solution set-ups:

a) x = (2.00 mol L¯¹) (2.35 L)
b) x = (0.415 mol L¯¹) (0.01600 L)
c) x = (0.500 mol L¯¹) (3.00 L)
d) x = (3.76 mol L¯¹) (6.20 L)

Comment: the technique used is this:

MV = moles of solute

This particular variation of the molarity equation occurs quite a bit in certain parts of the acid base unit.

Problem #18: Determine the grams of solute to prepare these solutions:

a) 0.289 liters of a 0.00300 M Cu(NO₃)₂ solution.
b) 16.00 milliliters of a 5.90-molar Pb(NO₃)₂ solution.
c) 508 mL of a 2.75-molar NaF solution.
d) 6.20 L of a 3.76-molar Na₂O solution.
e) 0.500 L of a 1.00 M KCl solution.
f) 4.35 L of a 3.50 M CaCl₂ solution.

Solution set-ups:

a) (0.00300 mol L¯¹) (0.289 L) = x / 187.56 g mol¯¹
b) (5.90 mol L¯¹) (0.01600 L) = x / 331.2 g mol¯¹
c) (2.75 mol L¯¹) (0.508 L) = x / 41.99 g mol¯¹
d) (3.76 mol L¯¹) (6.20 L) = x / 61.98 g mol¯¹
e) (1.00 mol L¯¹) (0.500 L) = x / 74.55 g mol¯¹
f) (3.50 mol L¯¹) (4.35 L) = x / 110.99 g mol¯¹

Problem #19: Determine the final volume of these solutions:

a) 4.67 moles of Li₂SO₃ dissolved to make a 3.89 M solution.
b) 4.907 moles of Al₂O₃ to make a 0.500 M solution.
c) 0.783 grams of Na₂CO₃ to make a 0.348 M solution.
d) 8.97 grams of (NH₄)₂CO₃ to make a 0.250-molar solution.
e) 48.00 grams of PbCl₂ to form a 5.0-molar solution.

Solution set-ups:

a) x = 4.67 mol / 3.89 mol L¯¹
b) x = 4.907 mol / 0.500 mol L¯¹
c) (0.348 mol L¯¹) (x) = 0.783 g / 105.99 g mol¯¹
d) (0.250 mol L¯¹) (x) = 8.97 g / 96.01 g mol¯¹
e) (5.00 mol L¯¹) (x) = 48.0 g / 278.1 g mol¯¹

Problem #20: A student placed 11.0 g of glucose (C₆H₁₂O₆) in a volumetric flask, added enough water to dissolve the glucose by swirling, then carefully added additional water until the 100. mL mark on the neck of the flask was reached. The flask was then shaken until the solution was uniform. A 20.0 mL sample of this glucose solution was diluted to 0.500L. How many grams of glucose are in 100. mL of the final solution?
Solution path #1:
1) Calculate molarity of first solution (produced by dissolving 11.0 g of glucose):

MV = grams / molar mass(x) (0.100 L) = 11.0 g / 180.155 g/mol
x = 0.610585 mol/L (I'll carry a few guard digits.)

2) Calculate molarity of second solution (produced by diluting the first solution):

M₁V₁ = M₂V₂(0.0200 L) (0.610585 mol/L) = (0.500 L) (x)
x = 0.0244234 mol/L

3) Determine grams of glucose in 100. mL of second solution:

MV = grams / molar mass(0.0244234 mol/L) (0.100 L) = x / 180.155 g/mol
x = 0.44 g

Solution path #2:
1) Calculate how much glucose you have in 20.0 mL of the first solution.

11.0 g is to 100. mL as x is to 20.0 mLCross-multiply and divide
100x = 11.0 times 20.0
x = 2.2 g

2) When you dilute the 20.0 mL sample to 500.0 mL, you have 2.2 g glucose in the solution.

2.2 g is to 500. mL as x is to 100. mLCross-multiply and divide
500x = 2.2 times 100
x = 0.44 g
In 100 mL of the final solution are 0.44 g glucose.

Problem #21: Commercial bleach solution contains 5.25% (by mass) of NaClO in water. It has a density of 1.08 g/mL. Caculate the molarity of this solution. (Hints: assume you have 1.00 L of solution; molar mass of NaClO 74.4 g/mol)
Solution:
1) Determine mass of 1.00 L of solution:

(1.08 g/mL) (1000 mL) = 1080 g

2) Determine mass of NaClO in 1080 g of solution:

(1080 g) (0.0525) = 56.7 g

3) Determine moles of NaClO:

56.7 g / 74.4 g/mol = 0.762 mol

4) Determine molarity of solution:

0.762 mol / 1.00 L = 0.762 M

Problem #22: What is the molality (and molarity) of a 20.0% by mass hydrochloric acid solution? The density of the solution is 1.0980 g/mL.
Solution: 1) Determine moles of HCl in 100.0 g of 20.0% solution.

20.0 % by mass means 20.0 g of HCl in 100.0 g of solution.20.0 g / 36.4609 g/mol = 0.548 mol

2) Determine molality:

0.548 mol / 0.100 kg = 5.48 m

3) Determine volume of 100.0 g of solution.

100.0 g / 1.0980 g/mL = 91.07468 mL

4) Determine molarity:

0.548 mol / 0.09107468 L = 6.02 m

Problem #23: 25.0 mL of 0.250 M KI, 25.0 mL of 0.100 K₂SO₄, and 15.0 mL of 0.100 M MgCl₂ were mixed together in a beaker. What are the molar concentrations of I¯, Cl¯ and K⁺ in the beaker?
Solution:
1) Calculate the total volume of the mixed solutions:

25.0 mL + 25.0 mL + 15.0 mL = 65.0 mL

2) Concentration of iodide ion:

M₁V₁ = M₂V₂(0.250 mol/L) (25.0 mL) = (x) (65.0 mL)
x = 0.09615 M
to three sig figs, 0.0962 M

3) Concentration of the chloride ion:

moles Cl¯ ---> (0.100 mol/L) (0.0150 L) (2 Cl¯ / 1 MgCl₂) = 0.00300 mol0.00300 mol / 0.065 L = 0.04615 M
to three sig figs, 0.0462 M

4) Concentration of the potassium ion:

moles K⁺ from KI ---> (0.250 mol/L) (0.0250 L) = 0.00625 mol
moles K⁺ from K₂SO₄ ---> (0.100 mol/L) (0.0250 L) (2 K⁺ / 1 K₂SO₄) = 0.00500 mol0.00625 mol + 0.00500 mol = 0.01125 mol
0.01125 mol / 0.0650 L = 0.173 M

Problem #24: Calculate the total concentration of all the ions in each of the following solutions:

a. 3.25 M NaCl
b. 1.75 M Ca(BrO₃)₂
c. 12.1 g of (NH₄)₂SO₃ in 615 mL in solution.

Solution:
1) the sodium chloride solution:

for every one NaCl that dissolves, two ions are produced (one Na⁺ and one Cl¯).the total concentration of all ions is this:
(3.25 mol/L) times (2 total ions / 1 NaCl formula unit) = 6.50 M

2) the calcium bromate solution:

three total ions are produced for every one Ca(BrO₃)₂ that dissolves (one Ca²⁺ and two BrO₃¯the total concentration of all ions is this:
(1.75 mol/L) times (3 total ions / 1 Ca(BrO₃)₂ formula unit) = 5.25 M

3) the ammonium sulfite solution:

calculate the concentration of (NH₄)₂SO₃:(x) (0.615 L) = 12.1 g / 116.1392 g/mol
x = 0.169407 M <--- I'll carry some guard digits
calculate the concentration of all ions:
(NH₄)₂SO₃ produces three ions for every one formula unit that dissolves.
0.169407 M times 3 = 0.508221 M
to three sig figs, 0.508 M

Problem #25: A solution of calcium bromide contains 20.0 g dm^-3. What is the molarity of the solution with respect to calcium bromide and bromine ions.
Solution:

MV = mass / molar mass(x) (1.00 L) = 20.0 g / 199.886 g/mol
x = 0.100 M
When CaBr₂ ionizes, two bromide ions are released for every one CaBr2 that dissolves. That leads to this:
[Br^-] = 0.200 M

Problem #26: What is the concentration of each type of ion in solution after 23.69 mL of 3.611 M NaOH is added to 29.10 mL of 0.8921 M H₂SO₄? Assume that the final volume is the sum of the original volumes.
Solution:
The answer requires you to know how NaOH and H₂SO₄ react. Here is the chemical equation:

H₂SO₄ + 2NaOH ---> Na₂SO₄ + 2H₂O

A key point is that two NaOH formula units are required for every one H₂SO₄
1) Calculate moles of NaOH and H₂SO₄:

moles NaOH ---> (3.611 mol/L) (0.02369 L) = 0.08554459 mol
moles H₂SO₄ ---> (0.8921 mol/L) (0.02910 L) = 0.02596011 mol

2) Determine how much NaOH remains after reacting with the H₂SO₄:

0.02596011 mol x 2 = 0.05192022 mol <--- moles of NaOH that react0.08554459 mol - 0.05192022 mol = 0.03362437 mol <--- moles of NaOH that remain

3) The above was required to determine the hydroxide ion concentration:

0.03362437 mol / 0.05279 L = 0.6369 M0.05279 L is the sum of the two solution volumes.

4) Determine the sodium ion concentration:

0.08554459 mol / 0.05279 L = 1.620 MThe sole source of sodium ion is from the NaOH.

5) Determine the sulfate ion concentration:

0.02596011 mol / 0.05279 L = 0.4918 MThe sole source of sulfate ion is from the H₂SO₄.

6) Determine the hydrogen ion concentration:

[H⁺] [OH¯] = 1.000 x 10^-14(x) (0.636945823) = 1.000 x 10^-14
x = 1.570 x 10^-14 M

Problem #27: Given 3.50 mL of sulfuric acid (98.0% w/w) calculate the number of mmols in the solution (density: 1.840 g/mL).
Solution:

3.50 mL times 1.840 g/mL = 6.44 g <--- mass of the 3.50 mL6.44 g times 0.980 = 6.3112 g <--- mass of H₂SO₄ in the solution
6.3112 g / 98.0768 g/mol = 0.06434957 mol
0.06434957 mol times (1000 mmol / 1 mol) = 64.3 mmol (to three sig figs)

Problem #28: Given 8.00 g of HBr calculate the volume (mL) of a 48.0% (w/w) solution. (MW HBr: 80.9119 g/mol, density: 1.49 g/mL). Then, calculate the molarity.
Solution:

8.00 g divided by 0.48 = 16.6667 g <--- total mass of the solution in which the HBr is 48% by mass16.6667 g divided by 1.49 g/mL = 11.18568 mL
to three sig figs, the volume of the solution is 11.2 mL
For the molarity, determine the moles of HBr:
8.00 g / 80.9119 g/mol = 0.098873 mol
0.098873 mol / 0.01118568 L = 8.84 M

Problem #29: A solution is made by dissolving 0.100 mol of NaCl in 4.90 mol of water. What is the mass % of NaCl?
Solution:
1) Convert moles to masses:

NaCl ---> 0.100 mol times 58.443 g/mol = 5.8443 gH₂O ---> 4.90 mol times 18.015 g/mol = 88.2735 g

2) Calculate mass percent of NaCl:

[5.8443 g / (5.8443 g + 88.2735 g)] * 100 = 6.21% (to three sig figs)

Problem #30: 2.00 L of HCl gas (measured at STP) is dissolved in water to give a total volume of 250. cm³ of solution. What is the molarity of this solution?
Solution using molar volume:

2.00 L divided by 22.414 L/mol = 0.0892299 mol of HCl0.0892299 mol / 0.250 L = 0.357 M (to three sig figs)

Solution using Ideal Gas Law:

PV = nRT(1.00 atm) (2.00 L) = (n) (0.08206 L atm / mol K) (273.15 K)
n = 0.0892272 mol
0.0892272 mol / 0.250 L = 0.357 M (to three sig figs)

If the volume of HCl gas was not at STP, you must use PV = nRT to calculate the moles. You cannot use molar volume since it is only true at STP.

Bonus Problem: How many milliliters of concentrated hydrochloric acid solution (36.0% HCl by mass, density = 1.18 g/mL) are required to produce 18.0 L of a solution that has a pH of 2.01?
Solution:
1) Get moles of hydrogen ion needed for the 18.0 L:

[H⁺] = 10^-pH = 10^-2.01 = 0.0097724 M0.0097724 mol/L) (18.0 L) = 0.1759032 mol of HCl required
Remember, HCl is a strong acid, dissociating 100% in solution

2) Determine molarity of 36.0% HCl:

Assume 100. g of solution present.36.0 g of that is HCl
100. g / 1.18 g/mL = 84.745763 mL
Use MV = mass / molar mass
(x) (0.084745763 L) = 36.0 g / 36.4609 g/mol
x = 11.6508 M

3) Volume of 11.6508 M acid needed to deliver 0.1759032 mol:

0.1759032 mol / 11.6508 mol/L = 0.01509795 L

15.1 mL (to three sig figs)

Molarity Dafination and Explanation

2016-05-27T02:29:00.001-07:00

Molarity Dafination and Explanation
Molarity Problems

As should be clear from its name, molarity involves moles. Boy, does it!
The molarity of a solution is calculated by taking the moles of solute and dividing by the liters of solution.

This is probably easiest to explain with examples.

Example #1: Suppose we had 1.00 mole of sucrose (it's about 342.3 grams) and proceeded to mix it into some water. It would dissolve and make sugar water. We keep adding water, dissolving and stirring until all the solid was gone. We then made sure that when everything was well-mixed, there was exactly 1.00 liter of solution.
What would be the molarity of this solution?

The answer is 1.00 mol/L. Notice that both the units of mol and L remain. Neither cancels.
A symbol for mol/L is often used. It is a capital M. So, writing 1.00 M for the answer is the correct way to do it.
Some textbooks make the M using italics and some put in a dash, like this: 1.00-M. When you handwrite it; a block capital M is just fine.
When you say it out loud, say this: "one point oh oh molar." You don't have to say the dash (if it's there). By the way, you sometimes see 1.00 M like this: 1.00-molar. A dash is usually used when you write the word 'molar.'
And never forget this: replace the M with mol/L when you do calculations. The M is the symbol for molarity, the mol/L is the unit used in calculations.

Example #2: Suppose you had 2.00 moles of solute dissolved into 1.00 L of solution. What's the molarity?

The answer is 2.00 M.
Notice that no mention of a specific substance is mentioned at all. The molarity would be the same. It doesn't matter if it is sucrose, sodium chloride or any other substance. One mole of sucrose or sodium chloride or anything else contains the same number of chemical units. And that number is 6.022 x 10²³ units, called Avogadro's Number.

Example #3: What is the molarity when 0.75 mol is dissolved in 2.50 L of solution?

The answer is 0.300 M.

Now, let's change from using moles to grams. This is much more common. After all, chemists use balances to weigh things and balances give grams, NOT moles.
Example #4: Suppose you had 58.44 grams of NaCl and you dissolved it in exactly 2.00 L of solution. What would be the molarity of the solution?
The solution to this problem involves two steps which will eventually be merged into one equation.

Step One: convert grams to moles.Step Two: divide moles by liters to get molality.

In the above problem, 58.44 grams/mol is the molar mass of NaCl. (There is the term "formula weight" and the term "molecular weight." There is a technical difference between them that isn't important right now. The term "molar mass" is a moe generic term.) To solve the problem:

Step One: dividing 58.44 grams by 58.44 grams/mol gives 1.00 mol.Step Two: dividing 1.00 mol by 2.00 L gives 0.500 mol/L (or 0.500 M).

Comment: remember that sometimes, a book will write out the word "molar," as in 0.500-molar.

Example #5: Calculate the molarity of 25.0 grams of KBr dissolved in 750.0 mL.

Example #6: 80.0 grams of glucose (C₆H₁₂O₆, mol. wt = 180. g/mol) is dissolved in enough water to make 1.00 L of solution. What is its molarity?

Notice how the phrase "of solution" keeps showing up. The molarity definition is based on the volume of the solution, NOT the volume of pure water used. For example, to say this:

"A one molar solution is prepared by adding one mole of solute to one liter of water."

is totally incorrect. It is "one liter of solution" not "one liter of water."
Be careful on this, especially when you get to molality.

The most typical molarity problem looks like this:
What is the molarity of "whatever" grams of "whatever" substance dissolved in "whatever" mL of solution.
To solve it, you convert grams to moles, then divide by the volume, like this:

The two steps just mentioned can be combined into one equation. Notice that moles is part of both equations, so one equation can be substituted into the other. Let's substitute the first into the second and rearrange just a bit to get this:

The M stands for molarity, the V for volume. In the second question, GMW has been substituted for molar mass. GMW stands for gram-molecular weight, which is a very common synonym for molar mass.

Example #7: When 2.00 grams of KMnO₄ (molec. wt = 158.0 g/mol) is dissolved into 100.0 mL of solution, what molarity results?

This next example is the most common type you'll see:
Example #8: How many grams of KMnO₄ are needed to make 500.0 mL of a 0.200 M solution?

Example #9: 10.0 g of acetic acid (CH₃COOH) is dissolved in 500.0 mL of solution. What molarity results?

Example #10: How many mL of solution will result when 15.0 g of H₂SO₄ is dissolved to make a 0.200 M solution?

Molality Problems

2016-05-27T02:26:00.001-07:00

Molality Problems

Problem #1: A solution of H₂SO₄ with a molal concentration of 8.010 m has a density of 1.354 g/mL. What is the molar concentration of this solution?
Solution:

8.010 m means 8.010 mol / 1 kg of solvent8.010 mol times 98.0768 g/mol = 785.6 g of solute
785.6 g + 1000 g = 1785.6 g total for solute and solvent in the 8.010 m solution.
1785.6 g divided by 1.354 g/mL = 1318.76 mL
8.01 moles / 1.31876 L = 6.0739 M
6.074 M (to four sig figs)

Problem #2: A sulfuric acid solution containing 571.4 g of H₂SO₄ per liter of solution has a density of 1.329 g/cm³. Calculate the molality of H₂SO₄ in this solution
Solution:

1 L of solution = 1000 mL = 1000 cm³1.329 g/cm³ times 1000 cm³ = 1329 g (the mass of the entire solution)
1329 g minus 571.4 g = 757.6 g = 0.7576 kg (the mass of water in the solution)
571.4 g / 98.0768 g/mol = 5.826 mol of H₂SO₄
5.826 mol / 0.7576 kg = 7.690 m

Problem #3: An aqueous solution is prepared by diluting 3.30 mL acetone (d = 0.789 g/mL) with water to a final volume of 75.0 mL. The density of the solution is 0.993 g/mL. What is the molarity, molality and mole fraction of acetone in this solution?
Solution:
1) Preliminary calculations:

mass of acetone: (3.30 mL) (0.789 g/mL) = 2.6037 g
moles of acetone: 2.6037 g / 58.0794 g/mol = 0.04483 mol <--- need to look up formula of acetone
mass of solution: (75.0 mL) (0.993 g/mL) = 74.475 g
mass of water in the solution: 74.475 g - 2.6037 g = 71.8713 g
moles of water: 71.8713 g / 18.015 g/mol = 3.9896 mol

2) Molarity:

0.04483 mol / 0.0750 L = 0.598 M

3) Molality:

0.04483 mol / 0.0718713 kg = 0.624 m

4) Mole fraction:

0.04483 mol / (0.04483 mol + 3.9896 mol) = 0.0111

Problem #4: Calculate the molality of 15.00 M HCl with a density of 1.0745 g/cm³
Solution:
1) Let us assume 1000. mL of solution are on hand. In that liter of 15-molar solution, there are:

15.00 mol/L times 1.000 L = 15.00 mole of HCl15.00 mol times 36.4609 g/mol = 546.9135 g of HCl

2) Use the density to get mass of solution

1000. mL times 1.0745 g/cm³ = 1074.5 g of solution1074.5 g minus 546.9135 g = 527.5865 g of water = 0.5275865 kg

3) Calculate molality:

15.00 mol / 0.5275865 kg = 28.43 m (to four sig figs)

Note: the mole fractions of water and HCl can also be calculated with the above data. There are 29.286 moles of water and 15.00 moles of HCl. You may work out the mole fractions on your own.

Problem #5: You are given 450.0 g of a 0.7500 molal solution of acetone dissolved in water. How many grams of acetone are in this amount of solution?
Solution:

0.7500 molal means 0.7500 mole of solute (the acetone) per 1000 g of watermass of acetone ---> 58.0794 g/mol times 0.7500 mol = 43.56 g
mass of solution ---> 1000 g + 43.56 g = 1043.56 g
43.56 is to 1043.56 as x is to 450
x = 18.78 g

Problem #6: A 0.391 m solution of the solute hexane dissolved in the solvent benzene is available. Calculate the mass (g) of the solution that must be taken to obtain 247 g of hexane (C₆H₁₄).
Solution:

0.391 mol times 86.1766 g/mol = 33.6950 g33.6950 g + 1000 g = 1033.6950 g
In other words, every 1033.6950 g of 0.391 m solution delivers 33.6950 g of hexane
33.6950 is to 1033.6950 as 247 is to x
x = 7577.46446 g
to three sig figs, 7.58 kg of solution

Problem #7: Calculate the mass of the solute C₆H₆ and the mass of the solvent tetrahydrofuran that should be added to prepare 1.63 kg of a solution that is 1.42 m.
Solution:

1.42 m means 1.42 mole of C₆H₆ in 1 kg of tetrahydrofuran1.42 mol times 78.1134 g/mol = 110.921 g
110.921 g + 1000 g = 1110.921 g
110.921 is to 1110.921 as x is to 1630
x = 162.75 g
To check, do this:
162.75 g / 78.1134 g/mol = 2.08351 mol
1630 g - 162.75 g = 1467.25 g
2.08351 mol / 1.46725 kg = 1.42 m

Problem #8: What is the molality of NaCl in an aqueous solution in which the mole fraction of NaCl is 0.100?
Solution:

A mole fraction of 0.100 for NaCl means the mole fraction of water is 0.900.Let us assume a solution is present made up of 0.100 mole of NaCl and 0.900 mole of water.
mass of water present ---> 0.900 mol times 18.015 g/mol = 16.2135 g
molality of solution ---> 0.100 mol / 0.0162135 kg = 6.1677 m
to three sig figs, 6.17 m

Problem #9: Calculate the molality (m) of a 7.55 kg sample of a solution of the solute CH₂Cl₂ (molar mass = 84.93 g/mol) dissolved in the solvent acetone (CH₃COH₃C) if the sample contains 929 g of methylene chloride
Solution:

mass solvent ---> 7550 g - 929 g = 6621 g = 6.621 kgmoles solute ---> 929 g/ 84.93 g/mol = 10.9384 mol
molality = 10.9384 mol / 6.621 kg = 1.65 m

Problem #10: What is the molality of a 3.75 M H₂SO₄ solution with a density of 1.230 g/mL?
Solution:
1) Determine mass of 1.00 L of solution:

1000 mL x 1.230 g/mL = 1230 g

2) Determine mass of 3.75 mol of H₂SO₄:

3.75 mol x 98.0768 g/mol = 367.788 g

3) Determine mass of solvent:

1230 - 367.788 = 862.212 g

4) Determine molality:

3.75 mol / 0.862212 kg = 4.35 molal (to three sig figs)

Problem #11: What is the molality of NaCl in an aqueous solution which is 4.20 molar? The density of the solution is 1.05 x 10³ g/L.
Solution:
1) Assume 1.00 L of the 4.20 M solution is present:

1.00 L of this solution contains 4.20 mole of NaCl.1.00 L times 1050 g/L = 1050 g of solution.

2) Determine mass of water in 1050 g of solution:

4.20 mol times 58.443 g/mol = 245.4606 g <--- mass of NaCl in solution1050 g - 245.4606 g = 804.5394 g

3) Calculate the molality:

4.20 mol / 0.8045394 kg = 5.22 m (to three sig figs)

Problem #12: Calculate the molarity of a 3.58 m aqueous RbCl solution with a density of 1.12 g/mL.
Solution:
1) 3.58 m means this:

3.58 mole of RbCl in 1000 g of water.

2) Determine total mass of solution:

3.58 mol times 120.921 g/mol = 432.89718 g1000 g + 432.89718 g = 1432.89718 g

3) Determine volume of solution:

1432.89718 g / 1.12 g/mL = 1279.37 mL

4) Determine molarity:

3.58 mol / 1.27937 L = 2.80 M

Here's another problem of this type.

Problem #13: Calculate the molality of a solution containing 16.5 g of naphthalene (C₁₀H₈) in 54.3 g benzene (C₆H₆).
Solution:

molality = moles of naphthalene / kilograms of benzene(16.5 g / 128.1732 g/mol) / 0.0543 kg = 2.37 m

Problem #14: What is the molality of a solution consisting of 1.34 mL of carbon tetrachloride (CCl₄, density= 1.59 g/mL) in 65.0 mL of methylene chloride (CH₂Cl₂, density = 1.33 g/mL)?
Solution:
1) Moles CCl₄:

1.34 mL times 1.59 g/mL = 2.1306 g2.1306 g / 153.823 g/mol = 0.013851 mol

2) Mass of the methylene chloride:

65.0 mL times 1.33 g/mL = 86.45 g = 0.08645 kg

3) Molality:

0.013851 mol / 0.08645 kg = 0.160 m (to three sig figs)

Problem #15: Determine concentration of a solution that contains 825 mg of Na₂HPO₄ dissolved in 450.0 mL of water in (a) molarity, (b) molality, (c) mole fraction, (d) mass %, and (e) ppm. Assume the density of the solution is the same as water (1.00 g/mL). Assume no volume change upon the addition of the solute.
Solution:
1) Molarity:

MV = mass / molar mass(x) (0.4500 L) = 0.825 g / 141.9579 g/mol
x = 0.0129 M

2) Molality:

0.825 g / 141.9579 g/mol = 0.00581158 mol0.00581158 mol / 0.4500 kg = 0.0129 m

3) Mole fraction:

Na₂HPO₄ ---> 0.825 g / 141.9579 g/mol = 0.00581158 mol
H₂O ---> 450.0 g / 18.015 g/mol = 24.97918401 molmole fraction of the water ---> 24.97918401 mol / (24.97918401 mol + 0.00581158 mol) = 0.9998

4) Mass percent:

water ---> (450 g / 450.825 g) * 100 = 99.8%

5) ppm:

ppm means the number of grams of solute per 1,000,000 grams of solution0.825 is to 450.825 as x is to 1,000,000

x = 1830 ppm

Molality Defination and Explanation With Example

2016-05-27T02:25:00.000-07:00

Molality Defination and Explanation With Example

Molality Problems

As is clear from its name, molality involves moles. Boy, does it!
The molality of a solution is calculated by taking the moles of solute and dividing by the kilograms of solvent.

This is probably easiest to explain with examples.

Example #1: Suppose we had 1.00 mole of sucrose (it's about 342.3 grams) and proceeded to mix it into exactly 1.00 liter water. It would dissolve and make sugar water. We keep adding water, dissolving and stirring until all the solid was gone. We then made sure everything was well-mixed.
What would be the molality of this solution? Notice that my one liter of water weighs 1000 grams (density of water = 1.00 g / mL and 1000 mL of water in a liter). 1000 g is 1.00 kg, so:

The answer is 1.00 mol/kg. Notice that both the units of mol and kg remain. Neither cancels.
A symbol for mol/kg is often used. It is a lower-case m and is often in italics, m. Some textbooks also put in a dash, like this: 1.00-m. However, if you write 1.00 m for the answer, without the italics, then that usually is correct because the context calls for a molality. Having said that, however, be aware that often m is used for mass, so be careful. (A lower-case m is also used for meter, but the context should be clear that m means molality.) Maybe including the dash would be wise if there might be a potential misunderstanding
When you say it out loud, say this: "one point oh oh molal." You don't have to say the dash.
And never forget this: replace the m with mol/kg when you do calculations. The m is a symbol that stands for mol/kg. It is not the actual unit.

Example #2: Suppose you had 2.00 moles of solute dissolved into 1.00 L of solvent. What's the molality?

The answer is 2.00 m.
Notice that no mention of a specific substance is mentioned at all. The molarity would be the same. It doesn't matter if it is sucrose, sodium chloride or any other substance. One mole of anything contains 6.022 x 10²³ units.

Example #3: What is the molality when 0.75 mol is dissolved in 2.50 L of solvent?

The answer is 0.300 m.

Now, let's change from using moles to grams. This is much more common. After all, chemists use balances to weigh things and balances give grams, NOT moles.
Example #4: Suppose you had 58.44 grams of NaCl and you dissolved it in exactly 2.00 kg of pure water (the solvent). What would be the molality of the solution?
The solution to this problem involves two steps.

Step One: convert grams to moles.Step Two: divide moles by kg of solvent to get molality.

In the above problem, 58.44 grams/mol is the molar mass of NaCl.

Step One: 58.44 g / 58.44 gr/mol = 1.00 mol.Step Two: 1.00 mol / 2.00 kg = 0.500 mol/kg (or 0.500 m).

Sometimes, a book will write out the word "molal," as in 0.500-molal.

Example #5: Calculate the molality of 25.0 grams of KBr dissolved in 750.0 mL pure water.

Example #6: 80.0 grams of glucose (C₆H₁₂O₆, mol. wt = 180. g/mol) is dissolved in1.00 kg of water. Calculate the molality.

Example #7: Calcuate the molality when 75.0 grams of MgCl₂ is dissolved in 500.0 g of solvent.

Example #8: 100.0 grams of sucrose (C₁₂H₂₂O₁₁, mol. wt. = 342.3 g/mol) is dissolved in 1.50 L of water. What is the molality?

Example #9: 49.8 grams of KI is dissolved in 1.00 kg of solvent. What is the molality?

In the molarity tutorial the phrase "of solution" kept showing up. The molarity definition is based on the volume of the solution. This makes molarity a temperature-dependent definition. However, the molality definition does not have a volume in it and so is independent of any temperature changes. This will make molality a very useful concentration unit in the area of colligative properties.
Lastly, it is very common for students to confuse the two definitions of molarity and molality. The words differ by only one letter and sometimes that small difference is overlooked.

Dilution Problems

2016-05-27T02:20:00.001-07:00

Dilution Problems

Problem #1: If you dilute 175 mL of a 1.6 M solution of LiCl to 1.0 L, determine the new concentration of the solution.
Solution:

M₁V₁ = M₂V₂(1.6 mol/L) (175 mL) = (x) (1000 mL)
x = 0.28 M

Note that 1000 mL was used rather than 1.0 L. Remember to keep the volume units consistent.

Problem #2: You need to make 10.0 L of 1.2 M KNO₃. What molarity would the potassium nitrate solution need to be if you were to use only 2.5 L of it?
Solution:

M₁V₁ = M₂V₂(x) (2.5 L) = (1.2 mol/L) (10.0 L)
x = 4.8 M

Please note how I use the molarity unit, mol/L, in the calculation rather than the molarity symbol, M.

Problem #3: How many milliliters of 5.0 M copper(II) sulfate solution must be added to 160 mL of water to achieve a 0.30 M copper(II) sulfate solution?
Solution:

M₁V₁ = M₂V₂(5.00 mol/L) (x) = (0.3 mol/L) (160 + x)
5x = 48 + 0.3x
4.7x = 48
x = 10. mL (to two sig figs)

The solution to this problem assumes that the volumes are additive. That's the '160 + x' that is V₂.

Problem #4: What volume of 4.50 M HCl can be made by mixing 5.65 M HCl with 250.0 mL of 3.55 M HCl?
Solution:
Here is the first way to solve this problem:

M₁V₁ + M₂V₂ = M₃V₃(3.55) (0.250) + (5.65) (x) = (4.50) (0.250 + x)
Where x is volume of 5.65 M HCl that is added
(0.250 + x) is total resultant volume
0.8875 + 5.65x = 1.125 + 4.50 x
1.15x = 0.2375
x= 0.2065 L
Total amount of 4.50 M HCl is then (0.250 + 0.2065) = 0.4565 L
Total amount = 456.5 mL

Here is the second way to solve this problem:

Since the amount of 5.65 M added is not asked for, there is no need to solve for it.M₁V₁ + M₂V₂ = M₃V₃
(3.55) (250) + (5.65) (x - 250) = (4.50) (x)
That way, x is the answer you want, the final volume of the solution, rather than x being the amount of 5.65 M solution that is added.

Problem #5: A 40.0 mL volume of 1.80 M Fe(NO₃)₃ is mixed with 21.5 mL of 0.808M Fe(NO₃)₃ solution. Calculate the molar concentration of the final solution.
Solution:
Let's use a slightly different way to write the subscripts:

M₁V₁ + M₂V₂ = M₃V₃

There is no standard way to write the subscripts in problems of this type.
Substituting:

(1.80) (40.0) + (0.808) (21.5) = (M₃) (40.0 + 21.5)M₃ = 1.45 M

Problem #6: To 2.00 L of 0.445 M HCl, you add 3.88 L of a second HCl solution of an unknown concentration. The resulting solution is 0.974 M. Assuming the volumes are additive, calculate the molarity of the second HCl solution.
Solution #1:

M₁V₁ + M₂V₂ = M₃V₃(0.445) (2.00) + (x) (3.88) = (0.974) (2.00 + 3.88)
x = 0.125 M

Solution #2:
1) Calculate moles HCl in 0.445 M solution:

(0.445 mol/L) (2.00 L) = 0.890 moles

2) Set up expression for moles of HCl in second solution:

(x) (3.88 L) = moles HCl in unknown solution

3) Calculate moles of HCl in final solution:

(0.974 mol/L) (5.88 L) = 5.73 moles

4) Moles of HCl in two mixed solutions = moles of HCl in final solution:

0.890 moles + [(x) (3.88 L)] = 5.73 molesx = 1.25 M (to three sig figs)

Problem #7: To what volume should you dilute 133 mL of an 7.90 M CuCl₂ solution so that 51.5 mL of the diluted solution contains 4.49 g CuCl₂?
Solution:
1) Find moles:

(4.49g CuCl₂) (1 mole CuCl₂ / 134.45 grams) = 0.033395 moles CuCl₂

2) Find the molarity of the 51.5 mL of the diluted solution that contains 4.49g CuCl₂:

(0.033395 moles CuCl₂) / (0.0515 liters) = 0.648 M

3) Use the dilution formula:

M₁V₁ = M₂V₂(7.90 M) (133 mL) = (0.648 M) (V₂)
V₂ = 1620 mL

You should dilute the 133 mL of an 7.90 M CuCl₂ solution to 1620 mL.

Problem #8: If volumes are additive and 95.0 mL of 0.55 M KBr is mixed with 165.0 of a BaBr₂ solution to give a new solution in which [Br¯] is 0.65 M, what is the concentration of the BaBr₂ used to make the new solution?
Solution:

moles of Br¯ from KBr: (0.55 mol/L) (0.095 L) = 0.05225 molmoles of Br¯ in final solution: (0.65 mol/L) (0.260 L) = 0.169 mol
moles Br¯ provided by the BaBr₂ solution: 0.169 - 0.05225 = 0.11675 mol
BaBr₂ provides two Br¯ per formula unit so (0.11675 divided by 2) moles of BaBr₂ are required for 0.11675 moles of Br¯ in the solution.
molarity of BaBr₂ solution: 0.058375 mol / 0.165 L = 0.35 M

Problem #9: 1.00 L of a solution is prepared by dissolving 125.6 g of NaF in it. If I took 180 mL of that solution and diluted it to 500 mL, determine the molarity of the resulting solution.
Solution:
1) Calculate moles of NaF:

125.6 g / 41.9 g/mol = 3.00 mol

2) Calculate moles in 180 mL of resulting solution:

3.00 mol in 1000 mL so 3 x (180/1000) = 0.54 mol in 180 mL

3) Calculate molarity of diluted solution:

0.54 mol / 0.50 L = 1.08 mol/L = 1.08 M

Problem #10: What is the molar concentration of chloride ions in a solution prepared by mixing 100.0 mL of 2.0 M KCl with 50.0 mL of a 1.50 M CaCl₂solution?
(Warning: there's a complication in the solution. It has to do with the CaCl₂.)
Solution #1:
1) Get total moles of chloride:

KCl ⇒ (2.00 mol/L) (0.100 L) = 0.200 mol of chloride ionCaCl₂ ⇒ (1.50 mol/L) (0.0500 L) (2 ions / 1 formula unit) = 0.150 mol of chloride ion.
The '2 ions / 1 formula unit' is the problem child. The solution is 1.50 M in calcium chloride, but 3.00 M in just chloride ion.
total moles = 0.200 mol + 0.150 mol = 0.350 mol

2) Get chloride molarity:

0.350 mol / 0.150 L = 2.33 M

Solution #2:
Suppose you really wanted to use this equation:

M₁V₁ + M₂V₂ = M₃V₃

Set it up like this:

(2.00 mol/L) (0.100 L) + (3.00 mol/L) (0.0500 L) = (M₃) (0.150 L)Note that the CaCl₂ molarity is 3.00 because that is the molarity of the solution from the point-of-view of the chloride ion

Problem #11: 46.2 mL of a 0.568 M Ca(NO₃)₂ solution is mixed with 80.5 mL of 1.396 M calcium nitrate solution. What is the nitrate ion concentration?
Solution:
1) Determine total moles of nitrate in solution:

(0.568 mol/L) (0.0462 L) (2 nitrate / formula unit) = 0.0524832 mol(1.396 mol/L) (0.0805 L) (2 nitrate / formula unit) = 0.224756 mol
0.0524832 mol + 0.224756 mol = 0.2772392 mol (of nitrate)

2) Determine nitrate concentration:

0.2772392 mol / 0.1267 L = 2.19 M (to three sig figs)

Problem #12: 66.8 mL of a 0.235 M solution of Ca(NO₃)₂ is mixed with 87.2 mL of a 0.450 solution of KNO₃. What is the nitrate ion concentration?
Solution:
1) Determine total moles of nitrate in solution:

(0.235 mol/L) (0.0668 L) (2 nitrate / formula unit) = 0.031396 mol(0.450 mol/L) (0.0872 L) (1 nitrate / formula unit) = 0.03924 mol
0.031396 mol + 0.03924 mol = 0.070636 mol (of nitrate)

2) Determine nitrate concentration:

0.070636 mol / 0.154 L = 0.459 M (to three sig figs)

Problem #13: 15.0 mL of 0.309 M Na₂SO₄ and 35.6 mL of 0.200 M KCl are mixed. Determine the following concentrations: [Na⁺]; [SO₄^2-]; [Cl^-]
Solution:
Comment: notice that no ion is in both solutions. That means that this problem is essentially three dilutions.
1) moles of each ion:

sodium: (0.309 mol/L) (0.0150 L) (2 sodium / formula unit) = 0.00927 molsulfate: (0.309 mol/L) (0.0150 L) (1 sulfate / formula unit) = 0.004635 mol
chloride: (0.200 mol/L) (0.0356 L) (1 chloride / formula unit) = 0.00712 mol

2) new molarities

sodium: 0.00927 mol / 0.0506 L = 0.183 Msulfate: 0.004635 mol / 0.0506 L = 0.0916 M
chloride: 0.00712 mol / 0.0506 L = 0.141 M

Problem #14: 3.50 g of NaCl is dissolved in 41.5 mL of 0.484 M CaCl₂ solution. Determine the chloride ion concentration.
Solution:
1) total moles of chloride:

from NaCl: (3.50 g / 58.443 g/mol (1 chloride / formula unit) = 0.0598874 molfrom CaCl₂: (0.484 mol/L) (0.0415 L) (2 chloride / formula unit) = 0.040172 mol
0.0598874 mol + 0.040172 mol = 0.1000594 mol

2) determine [Cl^-]:

0.1000594 mol / 0.0415 L = 2.41 M (to three sig figs)

Notice that I assumed the addition of the solid NaCl caused no volume change. There probably was some volume change, but, in the context of the problem, it was negligible.

Problem #15: How many milliliters of 1.5 M AlCl₃ must be used to make 70.0 mL of a solution that has a concentration of 0.21 M Cl¯?
Solution #1:

Think of the 1.5 M solution of AlCl₃ as being 4.5 M in chloride ion. This is because there are three chlorides in solution for every one AlCl₃dissolved.Use M₁V₁ = M₂V₂:

(4.5 mol/L) (x) = (0.21 mol/L) (70.0 mL)x = 3.27 mL

Notice how the mol/L cancels out and mL is left as the unit on the answer. You do not have to convert to liters for this particular type of problem, you just need to have the volumes be the same unit.

Solution #2:

(1.5 mol/L) (x) = (0.21 mol/L) (70.0 mL)x = 9.8 mL

However, you MUST realize how this answer is incorrect. We are being asked only for the chloride ion concentration. The 9.8 mL answer would give us a solution that is 0.21 M in aluminum chloride, AlCl₃. The concentration of ONLY the chloride ion is three times larger than the AlCl₃ concentration.
To obtain the correct answer, we must divide the 9.8 mL by three to get 3.27 mL. Be careful on this because you might think you should mltiply by three. Oh no. In the solution, the three winds up in the denominator (note where the 4.5 goes in solution #1. 4.5 is 1.5 times 3).
Personally, the ChemTeam thinks solution #1 is the better one. However, you do see some presentations that use #2.

Problem #16: A 40% acid solution is mixed with a 75% acid solution to produce 140 liters of a 50% acid solution? How many liters of each acid solution was mixed?
Comment: this will play a role:

percent times volume gives amount of pure acid present.

Solution:
1) Let x = the volume (in liters) of 40% acid. Therefore:

the amount of pure acid in that solution = (0.40)(x)

2) Since L 40% acid + L 75% acid = 140, then

the volume of 75% acid = 140 - x

3) The amount of pure acid in this solution:

(0.75)(140 - x)

4) The last set up conerns the final solution of 50% acid:

(0.50)(140 L) = 70 L pure acid

5) We will solve this:

L acid from 40% solution + L acid from 75% solution = total acid in 50% solution

6) Substitute and solve:

0.40x + (0.75)(140 - x) = 700.40x + 105 - 0.75x = 70
-0.35x = -35
x = 100 L
140 - x = 40 L
100 L of the 40% solution is mixed with 40 L of the 75% solution to yield 140 L of the 50% solution.

Problem #17: Container A holds a solution that is 80% alcohol while container B holds a solution that is 20% alcohol. How many liters of the solution in container A are needed to produce 12 liters of a solution that is 60% alcohol?
Solution:
Let us use this equation:

C₁V₁ = C₂V₂where C stands for concentration. It's a more general way to write the dilution equation.
(80) (x) + (20) (12 - x) = (60) (12)
80x + 240 - 20x = 720
60x = 480
x = 8 L
Notice that x is the volume of the 80% solution while 12 - x is the volume of the 20% solution.

Problem #18: What quantity of a 45% acid solution must be mixed with a 20% acid solution to produce 800 mL of a 29.375% solution?
Solution:

let x = mL of 45% acid
let y = mL of 20% acidx + y must equal 800 mL, therefore y = 800 - x
(x mL) (% acidity) + (y mL) (% acidity) = (mixture mL) (mixture acidity)
(x) (0.45) + (800 - x) (0.20) = (800) (0.29875)
The rest is left to the reader. Notice how the above solution uses decimal equivalents (0.45) rather than percents (45%). It doesn't matter which you use, just as long as you are consistent. Here's the set up with percents:
(x) (45) + (800 - x) (20) = (800) (29.875)
You'll get the same answer as with decimal equivalents.

Problem #19: A 74.31 g sample of Ba(OH)₂ is dissolved in enough water to make 2.450 L of solution. How many mL of this solution must diluted with water in order to make 1.000 L of 0.1000 molar Ba(OH)₂.
Solution:
1) Determine molarity of first solution:

MV = mass / molar mass(x) (2.450 L) = 74.31 g / 171.3438 g/mol
x = 0.1770161 M (I'll carry some guard digits.)

2) Determine volume required for dilution:

M₁V₁ = M₂V₂(0.1770161 mol/L) (x) = (0.1000 mol/L) (1000 mL)
x = 564.9 mL (to four sig figs)

Problem #20: How many milligrams of MgI₂ must be added to 234.0 mL of 0.0720 M KI to produce a solution with [I¯] = 0.1000 M?
Solution:

moles I¯ already dissolved ---> (0.0720 mol/L) (0.2340 L) = 0.016848 mol (KI produces iodide ion in a 1:1 molar ratio)moles I¯ desired in solution ---> (0.1000 mol/L) (0.2340 L) = 0.02340 mol
moles I¯ needed ---> 0.02340 - 0.016848 = 0.006552 mol
Every MgI₂ dissolved yields two iodine ions, so:
moles MgI₂ required ---> 0.006552 / 2 = 0.003276 mol
mass MgI₂ required ---> 0.003276 mol times 278.105 g/mol = 0.911072 g
convert to mg ---> 0.911072 g = 911 mg (to 3 sig figs, based on the 0.0720 M)

Problem #21: Determine the percentage concentration of the solution prepared by mixing 150 g of 20% solution and 250 g of 40%.
Solution #1:

150 g solution x 20% solute = 30 g solute
250 g solution x 40% solute = 100 g solute400 g solution containing 130 g of solute
% solute = (g solute / g solution) x 100 = (130 / 400) x 100 = 32.5%

Solution #2:

C₁M₁ + C₂M₂ = C₃M₃where C = concentration and M = mass of solution
(20) (150) + (40) (250) = (x) (400)
x = 32.5%

Problem #22: What volume of 0.416 M Mg(NO₃)₂ should be added to 255 mL of 0.102 M KNO₃ to produce a solution with a concentration of 0.278 M NO₃¯ ions? Assume volumes are additive.
Solution:

We need to consider the 0.416M Mg(NO₃)₂ solution for the standpoint of just the nitrate ion, in which case the concentration is twice the 0.416 value. So, [NO₃¯] in the Mg(NO₃)₂ solution is 0.832 M.In like manner, the [NO₃¯] in 0.102 M KNO₃ is 0.102 M.
M₁V₁ + M₂V₂ = M₃V₃
(0.832 mol/L) (x) + (0.102 mol/L) (255 mL) = (0.278 mol/L) (255 + x)
0.832x + 26.01 = 70.89 + 0.278x
0.554x = 44.88
x = 81.01083 mL
to three sig figs, the answer is 81.0 mL

Problem #23: What concentration of ClO₃^- results when 603 mL of 0.761 M AgClO₃ is mixed with 921 mL of 0.277 M Mn(ClO₃)₂?
Solution:
1) AgClO₃ ionizes as follows:

AgClO₃ ---> Ag⁺ + ClO₃^-

2) moles AgClO₃ dissolved:

(0.761 mol/L) (0.603 L) = 0.458883 mol

3) From the chemical equation, one mole of AgClO₃ dissolving yields one mole of ClO₃^- in solution.

moles ClO₃^- = 0.458883 mol

4) Mn(ClO₃)₂ dissolves as follows:

Mn(ClO₃)₂ ---> Mn²⁺ + 2ClO₃^-

5) moles Mn(ClO₃)₂ dissolved:

(0.277 mol/L) (0.921 L) = 0.255117 mol

6) From the chemical equation, one mole of Mn(ClO₃)₂ dissolving yields two moles of ClO₃^- in solution.

moles ClO₃^- = 0.255117 mol x 2 = 0.510234 mol

7) total moles ClO₃^- in solution:

0.510234 mol + 0.458883 mol = 0.969117 mol

8) total volume of solution:

0.921 L + 0.603 L = 1.524 L

9) molarity of ClO₃^-:

0.969117 mol / 1.524 L = 0.636 M (to three sig figs)

Problem #24: Assuming the volumes are additive, what is the [NO₃^-] in a solution obtained by mixing 265 mL of 0.290 M KNO₃, 318 mL of 0.437 M Mg(NO₃)₂, and 774 mL of H₂O
Solution:
1) Moles nitrate ion from KNO₃:

(0.290 mol/L) (0.265 L) = 0.07685 molSince one nitrate ion is produced for every one KNO₃ that dissolves, 0.07685 mol of nitrate ion is produced.

2) Moles nitrate ion from Mg(NO₃)₂:

(0.437 mol/L) (0.318 L) = 0.138966 molSince two nitrate ions are produced for every one Mg(NO₃)₂ that dissolves,

0.138966 mol x 2 = 0.277932 mol
of nitrate ion is produced.

3) Total moles of nitrate produced:

0.138966 mol + 0.277932 mol = 0.416898 mol

4) Total volume of solution:

265 mL + 318 mL + 774 mL = 1357 mL = 1.357 L

5) Molarity of nitrate ion:

0.416898 mol / 1.357 L = 0.307 M (to three sig figs)

Problem #25: Describe the preparation of 900.0 mL of 3.00 M HNO₃ from the commercial reagent that is 70.5% HNO₃ (w/w) and has a specific gravity of 1.42.
Solution:

Let us start with 1.00 kg of the 70.5% reagent. From the 70.5%, we know that 705 g of HNO₃ are present. From the specific gravity of 1.42, we know the density to be 1.42 g/mL, so the volume of the 1.00 kg of solution is this:1000 g / 1.42 g/mL = 704.2 mL
moles of HNO₃ ---> 705 g / 63.0119 g/mol = 11.19 mol
molarity of the 70.5% solution ---> 11.19 mol / 0.7042 L = 15.89 M
Now, use M₁V₁ = M₂V₂
(3.00 mol/L) (900 mL) = (15.89 mol/L) (x)
x = 169.92 mL
To three sig figs, this is 170. mL
Diluting 170. mL of the 70.5% solution to a final volume of 900.0 mL will create the 3.00-molar solution we desire.

Bonus Problem: What volume of a concentrated HCl solution, which is 36.0% HCl by mass and has a density of 1.179 g/mL, should be used to make 4.75 L of an HCl solution with a pH of 1.6?
Solution:
1) pH 1.6 leads to this concentration:

10^-1.6 = 0.02511 M H⁺Since HCl is a strong acid, this is also the concentration of the HCl desired.

2) Determine molarity of concentrated HCl:

1.00 L of the HCl ---> 1000 mL times 1.179 g/mL = 1179 g
1179 g times 0.36 = 424.44 g of HCl
(424.44 g/36.46 g/mole) / 1.00 L = 11.64 M.

3) Determine volume of concentrated HCl to be diluted:

M₁V₁ = M₂V₂(11.64 mol/L) (x) = (0.02511 mol/L) (4.75 L)
x = 0.010247 L
x = 10.25 mL

When I copied this problem from Yahoo Answers, the solution was stated differently than I put it above. I offer it here for your study:

And you should dilute this concentrated solution 11.64/0.02511 fold = 463.56 fold.4750 mL / 463.56 = 10.25 mL. So use 10.25 mL of the concentrated HCl and fill up with water to 4.75 L.

Problem #26:
How many milliliters of 2.00 M copper(II) sulfate solution must be added to 165 mL of water to achieve a 0.300 M copper(II) sulfate solution?
Solution:

We will assume that volumes are additive.M₁V₁ = M₂V₂
(2.00 mol/L) (x) = (0.300 mol/L) (165 + x)
2x = 49.5 + 0.3x
1.7x = 49.5
x = 29.1 mL

Problem #27: Calculate the volume of solution prepared by diluting 6.929 mL of 3.555 M solution to 0.8229 M.
Solution:

Use this equation: M₁V₁ = M₂V₂(3.555 mol/L) (6.929 mL) = (0.8229 mol/L) (x)
x = 29.93388 mL
To four sig figs, this is 29.93 mL

Problem #28: Calculate the concentration of formaldehyde (CH₂O) in a solution prepared by mixing 125 mL of 6.13 M CH₂O and 175 mL of 4.34 M CH₂O and diluting the mixture to 500.0 mL.
Solution:
1) Determine the total moles of CH₂O in the two solutions being mixed:

(6.13 mol/L) (0.125 L) = 0.76625 mol
(4.34 mol/L) (0.175 L) = 0.7595 mol0.76625 mol + 0.7595 mol = 1.52575 mol

2) The total moles are diluted to a final volume of 500.0 mL:

1.52575 mol / 0.5000 L = 3.05 M (to three sig figs)

Although you are not asked, you can easily determine the molarity of the two formaldehyde solutions after they are mixed, but before they are diluted to the final volume of 500.0 mL:

1.52575 mol / 0.300 L = 5.08 M (to three sig figs)

You could have also used this:

M₁V₁ + M₂V₂ = M₃V₃

and solved for M₃.

Problem #29: Calculate the following quantity: volume of 2.48 M calcium chloride that must be diluted with water to prepare 356.0 mL of a 0.0586 chloride ion solution. (give answer in mL)
Solution:

The key to solving this problem is to know the formula of calcium chloride is CaCl₂. That means that, while the solution is 2.48 M in CaCl₂, it is 4.96 M from the perspective of just the chloride.We use M₁V₁ = M₂V₂ to solve this problem.
(4.96 mol/L) (x) = (0.0586 mol/L) (356.0 mL)
x = 4.205968 mL
to three sig figs, this is 4.20 mL

Problem #30: The concentration of muriatic acid is 11.7 M. A diluted solution of 3.50 M is prepared. How many milliliters of 3.50 M muriatic acid solution contains 35.7 g of HCl? (give answer in mL)
Solution:

Use this: MV = mass / molar mass(3.50 mol/L) (x) = 35.7 g / 36.4609 g/mol
x = 0.27975 L
to three sig figs, 280. mL

Problem #31: Determine the mass (g) of calcium nitrate in each milliliter of a solution prepared by diluting 56.0 mL of 0.705 M calcium nitrate to a final volume of 0.100 L
Solution:
1) use M₁V₁ = M₂V₂:

(0.705 mol/L) (0.0560 L) = (x) (0.100 L)x = 0.3948 M

2) moles of Ca(NO₃)₂ in 1 mL:

0.3948 mol/L = 0.3948 mol / 1000 mL = 0.0003948 mol/mL

3) Convert moles to grams:

0.0003948 mol/mL times 164.086 g/mol = 0.0648 g/mL

Problem #32: Concentrated sulfuric acid is 98.0% H₂SO₄ by mass and has a density of 1.84 g/mL. Determine the volume of acid required to make 1.00 L of 0.100 M H₂SO₄ solution.
Solution:

Assume 100. g of the solution is present. This means 98.0 g of H₂SO₄ are present.98.0 g / 98.0768 g/mol = 0.999217 mol
100. g / 1.84 g/mL = 54.34783 mL <--- volume of the 100. g of solution
molarity of the 98.0% solution:
0.999217 mol / 0.05434783 L = 18.3856 M
Now, use M₁V₁ = M₂V₂
(18.3856 mol/L) (x) = (0.1 mol/L) (1 L)
x = 0.005439 L = 5.44 mL

Problem #33a: What is the [NO₃¯] in 200. mL of 0.350 M Al(NO₃)₃?
Problem #33b: What is the [NO₃¯] in the solution above after adding 200.0 mL of 0.150 M Ca(NO₃)₂
Solution:

For every one formula unit of Al(NO₃)₃ that dissolves, three nitrate ions are released to the solution.0.350 M times 3 = 1.05 M in nitrate. The 200. mL is not needed.
For the second part, the 200. mL is needed.
(0.200 L) (1.05 mol/L) = 0.210 mol of nitrate
(0.200 L) (0.300 mol/L) = 0.060 mol of nitrate
0.210 + 0.060 = 0.270 mol of nitrate in 0.400 L of total volume
0.270 mol / 0.400 L = 0.675 M
I got the 0.300 by doing 0.150 times 2 since each Ca(NO₃)₂ delivers 2 nitrates to the solution.

Problem #34: What volume of a 15.0% by mass NaOH solution, has a density of 1.116 g/mL, should be used to make 5.30 L of an NaOH solution with a pH of 11.00?
Solution:
1) A pH of 11.00 means a pOH of 3.00 which means a hydroxide concentration of 0.0010 M. The total moles of hydroxide is this:

(0.0010 mol/L) (5.30 L) = 0.0053 mol

2) The mass of NaOH needed is this:

40.0 g/mol times 0.0053 mol = 0.212 g

3) The mass of 15/0% solution required is found by ratio and proportion:

0.212 g is to x as 15 is to 100x = 1.41333 g <--- kept a couple extra digits

4) The volume of NaOH solution required is this:

1.41333 g / 1.116 g/mL = 1.27 mL1.27 mL of 15.0%(w/w) NaOH solution, when diluted to 5.30 L of solution, has a pH equal to 1.00.

Problem #35: If a solution of MgCl₂ is 1/8 M, what will its concentration be if it is diluted by 27%?
Solution:

M₁V₁ = M₂V₂(0.125) (100 mL) = (x) (127 mL)
x = 0.098 M

Bonus Problem: What volume of a concentrated HCl solution, which is 36.0% HCl by mass and has a density of 1.179 g/mL, should be used to make 4.30 L of an HCl solution with a pH of 1.86?
Solution:

From the pH, you know that in the final solution, [H⁺] = 10^-1.86 = 0.0138 M (Quite properly, the pH has only 2 significant figures. The "1" is a place-holder digit. So, formally, you should say that the solution will have [H⁺] = 0.014 M, but we'll keep the third digit anyway...)Since HCl is a strong acid, we know that the final [HCl] will equal 0.0138 M.
Now, the problem is to calculate the molarity of the original HCl solution, and then use M₁V₁ = M₂V₂ to get your final answer.
36.0% by mass means that 100 g of the HCl solution will contain 36.0 grams of HCl.
36.0 g HCl / 36.45 g/mol = 0.988 moles HCl in 100 g solution.
The volume of 100 g of solution is: 100 g / 1.179 g/mL = 84.8 mL or 0.0848 L. So, the molarity of the original HCl solution is:
0.988 mol / 0.0848 L = 11.6 M
Now, use M₁V₁ = M₂V₂:
(11.6 mol/L) (V₁) = (0.0138 mol/L) (4.30 L)
V₁ = 5.11 x 10^-3 L = 5.11 mL

Dilution: Definition and Calculations

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Dilution: Definition and Calculations

To dilute a solution means to add more solvent without the addition of more solute. Of course, the resulting solution is thoroughly mixed so as to ensure that all parts of the solution are identical.
The fact that the solute amount stays constant allows us to develop calculation techniques.
First, we write:

moles solute before dilution = moles solute after dilution

From rearranging the equation that defines molarity, we know that the moles of solute equals the molarity times the volume. (Calculating the moles of solute from molarity times volume will be very useful in other areas of chemistry, particularly acid base. Remember that the volume must be in liters.)
So we can substitute MV (molarity times volume) into the above equation, like this:

M₁V₁ = M₂V₂

The "sub one" refers to the situation before dilution and the "sub two" refers to after dilution.
This equation does not have an official name like Boyle's Law, so we will just call it the dilution equation.

Example #1: 53.4 mL of a 1.50 M solution of NaCl is on hand, but you need some 0.800 M solution. How many mL of 0.800 M can you make?
Solution:
Using the dilution equation, we write:

(1.50 mol/L) (53.4 mL) = (0.800 mol/L) (x)x = 100. mL

Notice that the volumes need not be converted to liters. Any old volume measurement is fine, just so long as the same one is used on each side. (However, as mentioned above, if you are calculating how many moles of solute are present, you need to have the volume in liters.)

Example #2: 100.0 mL of 2.500 M KBr solution is on hand. You need 0.5500 M. What is the final volume of solution which results?
Placing the proper values into the dilution equation gives:

(2.500 mol/L) (100.0 mL) = (0.5500 mol/L) (x)x = 454.5454545 mL (oops, my fingers got stuck typing.) (Bad attempt at humor, really bad!)
x = 454.5 mL

Sometimes the problem might ask how much more water must be added. In this last case, the answer is 454.5 - 100.0 = 354.5 mL.
Go ahead and answer the question, if your teacher asks it, but it is bad technique in the lab to just measure out the "proper" amount of water to add and then add it. This is because the volumes (the soltion and the diluting water) are not necessarily additive. The only volume of importance is the final solution's volume. You add enough water to get to the final solution volume without caring how much the actual volume of water you added is.

Example #3: A stock solution of 1.00 M NaCl is available. How many milliliters are needed to make 100.0 mL of 0.750 M

(0.750 mol/L) (100.0 mL) = (1.00 mol/L) (x)x = 75.0 mL

Example #4: What volume of 0.250 M KCl is needed to make 100.0 mL of 0.100 M solution?

(0.100 mol/L) (100.0 mL) = (0.250 mol/L) (x)Please go ahead and solve for x.

Example #5: 2.00 L of 0.800 M NaNO₃ must be prepared from a solution known to be 1.50 M in concentration. How many mL are required?

(1.50 mol/L) (x) = (0.800 mol/L) (2.00 L)x = 1.067 L
Divide the liters by 1000 to get mL, the answer is 1070 mL (notice that it is rounded off to three sig figs)

These next two are a bit harder and involve slightly more calculation than the discussion above. Here's a summary of the steps:

1) calculate total moles
2) calculate total volume
3) divide moles by volume to get molarity

You can also think of it this way:

M₁V₁ + M₂V₂ = M₃V₃

Where the '1' refers to one starting solution, '2' to the other starting solution and '3' refers to the mixed solution (hints: V₃ is the total volume after mixing and M₃ is almost always the unknown).

Example #6: Calculate the final concentration if 2.00 L of 3.00 M NaCl and 4.00 L of 1.50 M NaCl are mixed. Assume there is no volume contraction upon mixing.
Here are the two mole calculations:

x = (3.00 mol/L) (2.00 L)
x = (1.50 mol/L) (4.00 L)

I hope it is obvious that you add the two answers to get the total moles.
The total volume calculation is 2.00 + 4.00 = 6.00 L.
Divide total moles by total volume to get the final answer. The answer is 2.00 M.
Using M₁V₁ + M₂V₂ = M₃V₃, we have this:
(3.00 mol/L) (2.00 L) + (1.50 mol/L) (4.00 L) = (x) (6.00 L)
6.00 mol + 6.00 mol = (x) (6.00 L)
x = 12.0 mol / 6.00 L = 2.00 mol/L

Example #7: Calculate the final concentration if 2.00 L of 3.00 M NaCl, 4.00 L of 1.50 M NaCl and 4.00 L of water are mixed. Assume there is no volume contraction upon mixing.
The solution to this problem is almost exactly the same as 10a. The only "problem child" appears to be the 4.00 L of water. Hint: the water contributes to the final volume, but NOT to the total moles. The ChemTeam gets a final answer of 1.20 M in this problem.
Using M₁V₁ + M₂V₂ = M₃V₃, we have this:
(3.00 mol/L) (2.00 L) + (1.50 mol/L) (4.00 L) = (x) (10.0 L)
6.00 mol + 6.00 mol = (x) (10.0 L)
x = 12.0 mol / 10.0 L = 1.20 M

Percentage compositions

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Percentage compositions

Percentage composition is the percentage by mass of the atoms of an element present in one mole of a compound.

This may also be obtained from the molecular formula of the compound by using the relationship

Example: 13

Calculate the percentage composition of each of the elements present in pure calcium carbonate (Ca = 40, C = 12, O = 16).

Solution:

Mass of one mole of pure calcium carbonate (CaCO₃)

= Mass of calcium + Mass of carbon + Mass of oxygen

= (40 x 1) + (12 x 1) + (16 x 3)

= 40 + 12 + 48

= 100 g

Example: 14

Calculate the percentage loss of mass when sodium carbonate crystals are strongly heated (Na=23, C=12, O=16, H=1).

Solution:

Mass of crystals of Na₂SO₃.10H₂O = 286

Mass of water of crystallisation = 180

Loss of mass = 180

= 62.94%

Example: 15

Calculate the percentage composition of the elements present in pure calcium sulphate. (Ca = 40, S= 32, O =16).

Solution:

Mass of 1 mole of CaSO₄ = 40 + 32 + (16 x 4)

=40 + 32 + 64 = 136

Example: 16

Calculate the percentage composition of the elements present in sodium sulphate decahydrate (Na=23, S=32, O=16, H=1)

Solution:

Mass of 1 mole of Na₂SO₄.10H₂O

= (23 x 2) + (32 x 1) + (16 x 4) + 10 (1 x 2 + 16)

or, combining the oxygen atoms

= 46 + 32 + 64 + 20 + 160 = 322

Derivation of empirical and molecular formula from percentage composition

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Derivation of empirical and molecular formula from percentage composition

Derivation of empirical formula from percentage composition:

Empirical formula of a compound is a formula that shows the simplest whole number ratio between the atoms of the elements in the compound.

It can be calculated from the percentage composition of the elements present in a compound. The various steps involved in determining the empirical formula are:

1) Divide the percentage composition of the elements present in a compound by their relative atomic mass. The quotients will be the ratio of the number of atoms.

2) The quotients obtained are then divided by the smallest quotient. This gives the simplest ratio between the atoms.

3) After the simplest ratio is obtained, multiply by any convenient integer to give the simplest whole number ratio.

4) The empirical formula is derived by writing the symbol of the elements present in it, followed by the number of atoms of each, as subscript, to the right of the symbol.

Derivation of molecular formula from percentage composition:

Molecular Formula of a substance gives the names and number of atoms of the various elements present and the molecular mass of it.

To find the molecular formula from percentage composition we have to undertake the following steps:

1) First find the empirical formula, as illustrated earlier.

2) Then find the empirical formula mass: for this we have to add the atomic masses, of all the atoms in the empirical formula.

For e.g., assume the empirical formula of a compound is HO.

Empirical formula mass = Mass of 1 hydrogen atom + mass of 1 oxygen atom

= 1 + 16 = 17

3) The relative molecular mass or molar mass is obtained from experiment or from the vapour density relationship: molecular mass = 2 x vapour density.

In the example cited above let the molar mass be = 34.

4) We then have to find the ratio (n) of the relative molecular mass to the empirical formula mass. This ratio is calculated as

From the example above 'n'

5) The molecular formula is found from the above ratio by using the realtion:

Molecular formula = n x empirical formula

In the example, the relative molecular mass is found to be two times greater than the empirical formula mass,

the molecular formula = empirical formula x 2

= (HO) x 2 = H₂O₂

Problems

_{8. An organic compound weighing 0.2 g containing C, H and O gave on combustion 0.296 g CO₂ and 0.12 g H₂O. Its molecular mass is 180. Determine its empirical and molecular formulae.

Solution

Mass of the compound taken = 0.2 g
Mass of CO₂ formed = 0.296 g

Mass of H₂O formed = 0.12 g

Therefore, the percentage of O = 100 - (40.0 + 6.67) = 53.33(by difference)
Calculation of empirical formula

Therefore, empirical formula of the compound = CH₂O
Calculation of molecular formula

Empirical formula mass = (1 x 12 amu + 2 x 1 amu + 1 x 16 amu)
= 30 amu
Molecular mass = 180 amu
Hence,

Therefore, Molecular formulae of a compound = 6 x CH₂O = C₆H₁₂O₆
9. A compound (molecular mass 147) contains 49.0 % carbon and 2.72% hydrogen. 2.561 mg of the compound gave 5.00 mg of silver chloride in Carius estimation. Determine its molecular formula. Find its empirical formula also.

Solution

The problem is done in steps:
Calculation of the percentage of chlorine:
We know that, Cl = AgCl
35.5g (108 +35.5)= 143.5? 5 mg

Then,
Percentage of chlorine in the sample =

Empirical formula = C₃H₂Cl
Empirical formula mass = (3 x 12 amu + 1 amu + 1 x 35.5 amu)
= 73.5 amu
Molecular mass (given) = 147 amu
Hence,

Therefore,
Molecular formula = 2 x Empirical formula
= 2 x C₃H₂Cl
= C₆H₄Cl₂
10. About 0.45 g of an organic compound gave on combustion 0.792 g of CO₂ and 0.324 g of water 0.24 g of the same substance was Kjeldahlised and the ammonia evolved was absorbed in 50.0cm³ of N/4 H₂SO₄. The excess of the acid required 77.0 cm³ of N/10 NaOH for complete neutralization. Calculate the empirical formula of the acid.

Solution

Calculation of the percentages of different elements.

Percentage of nitrogen can be calculated thus,
Volume of H₂SO₄ taken = 50 cm³ N/4 of solution
Volume of NaOH required for excess acid = 77 cm³ N/10 of solution
To calculate volume of N/4 H₂SO₄ used for neutralization of NH₃:
N₁V₁ (acid left unused) = N₂V₂ (alkali used)

Volume of N/4 H₂SO₄ used fro neutralizing NH₃= 50 30.8 = 19.2 cm³
Now, 1000 cm³ of N/4 contain nitrogen = 14 g

Percentage of oxygen = 100 - (%C + %H + %N)
= 100 - (48 + 8 + 28) =16
Calculation of empirical formula

The simplest ratio of C : H : N : O is 4 : 8 : 2 : 1
Empirical formula of the compound is C₄H₈N₂O.}

Chemical calculation based on Limiting reagents

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Chemical calculation based on Limiting reagents

Key Concepts

The limiting reagent is the reactant that is completely used up during the chemical reaction.
The reactant that is in excess is the reactant that is not completely used up during the chemical reaction, that is, there is some of this reactant left over.

Deciding which reactants are the limiting reagents and the reactants in excess

Write the balanced chemical equation for the chemical reaction
Calculate the available moles of each reactant in the chemical reaction
Use the balanced chemical equation to determine the mole ratio of the reactants in the chemical reaction
Compare the available moles of each reactant to the moles required for complete reaction using the mole ratio
The limiting reagent is the reactant that will be completely used up during the chemical reaction. There will be some moles of the reactant in excess left over after the reaction has gone to completion.

Examples

Moles of reactants given

Find the limiting reagent and the reactant in excess when 0.5 moles of Zn react completely with 0.4 moles of HCl

Write the balanced chemical equation for the chemical reactionZn + 2HCl -----> ZnCl₂ + H₂
Calculate the available moles of each reactant in the chemical reaction

moles of Zn = 0.5 moles of HCl = 0.4
Use the balanced chemical equation to determine the mole ratio of the reactants in the chemical reaction

Zn : HCl Or HCl : Zn

1 : 2 1 : ½
Compare the available moles of each reactant to the moles required for complete reaction using the mole ratioIf all of the 0.5 moles of Zn were to be used in the reaction it would require
2 x 0.5 = 1.0 moles of HCl for the reaction to go to completion.
There are only 0.4 moles of HCl available which is less than the required 1.0 moles.
If all of the 0.4 moles of HCl were to be used in the reaction it would require
½ x 0.4 = 0.2 moles Zn.
There are 0.5 moles of Zn available which is more than the required 0.2 moles.
The limiting reagent is the reactant that will be completely used up during the chemical reaction.
There will be some moles of the reactant in excess left over after the reaction has gone to completion.The limiting reagent is HCl,
all of the 0.4 moles of HCl will be used up when this reaction goes to completion.
The reactant in excess is Zn,
when the reaction has gone to completion there will be
0.5 - 0.2 = 0.3 moles of Zn left over.

Masses of reactants given

Find the limiting reagent and the reactant in excess when 1.5g of CaCO₃ react completely with 0.73g of HCl

Write the balanced chemical equation for the chemical reactionCaCO₃ + 2HCl -----> CaCl₂ + CO₂ + H₂O

Calculate the available moles of each reactant in the chemical reaction

moles of CaCO₃ = mass ÷ MM
mass = 1.5g
MM = 40.08 + 12.01 + (3 x 16.00)
= 100.09 g/molemoles of CaCO₃ = 1.5 ÷ 100.09
= 0.015 mol

moles of HCl = mass ÷ MM
mass = 0.73g
MM = 1.008 + 35.45
= 36.458g/molmoles HCl = 0.73 ÷ 36.458
= 0.02 mol

Use the balanced chemical equation to determine the mole ratio of the reactants in the chemical reaction

CaCO₃ : HCl Or HCl : CaCO₃

1 : 2 1 : ½
Compare the available moles of each reactant to the moles required for complete reaction using the mole ratioIf all of the 0.015 moles of CaCO₃ were to be used in the reaction it would require
2 x 0.015 = 0.03 moles of HCl for the reaction to go to completion.
There are only 0.02 moles of HCl available which is less than the required 0.03 moles.
If all of the 0.02 moles of HCl were to be used in the reaction it would require
½ x 0.02 = 0.01 moles of CaCO₃.
There are 0.015 moles of CaCO₃ available which is more than the required 0.01 moles.
The limiting reagent is the reactant that will be completely used up during the chemical reaction.
There will be some moles of the reactant in excess left over after the reaction has gone to completion.The limiting reagent is HCl,
all of the 0.02 moles of HCl will be used up when this reaction goes to completion.
The reactant in excess is CaCO₃,
when the reaction has gone to completion there will be
0.015 - 0.01 = 0.005 moles of CaCO₃ left over.

Concentration and volume of solutions given

Find the limiting reagent and the reactant in excess when 100mL of 0.2 NaOH react completely with 50mL of 0.5M H₂SO₄

Write the balanced chemical equation for the chemical reaction2NaOH + H₂SO₄ -----> Na₂SO₄ + 2H₂O
Calculate the available moles of each reactant in the chemical reaction

moles of NaOH = M x V
M = 0.2M
V = 100mL
= 100 x 10^-3Lmoles of NaOH = 0.2 x 100 x 10^-3
= 0.02 mol moles of H₂SO₄ = M x V
M = 0.5M
V = 50mL
= 50 x 10^-3Lmoles H₂SO₄ = 0.5 x 50 x 10^-3
= 0.025mol
Use the balanced chemical equation to determine the mole ratio of the reactants in the chemical reaction

NaOH : H₂SO₄ Or H₂SO₄ : NaOH

1 : ½ 1 : 2
Compare the available moles of each reactant to the moles required for complete reaction using the mole ratioIf all of the 0.02 moles of NaOH were to be used in the reaction it would require
½ x 0.02 = 0.01 moles of H₂SO₄ for the reaction to go to completion.
There are 0.025 moles of H₂SO₄ available which is more than the required 0.01 moles.
If all of the 0.025 moles of H₂SO₄ were to be used in the reaction it would require
2 x 0.025 = 0.05 moles of NaOH.
There are only 0.02 moles of NaOH available which is less than the required 0.05 moles.
The limiting reagent is the reactant that will be completely used up during the chemical reaction.
There will be some moles of the reactant in excess left over after the reaction has gone to completion.The limiting reagent is NaOH,
all of the 0.02 moles of NaOH will be used up when this reaction goes to completion.
The reactant in excess is H₂SO₄,
when the reaction has gone to completion there will be
0.025 - 0.01 = 0.015 moles of H₂SO₄ left over.

Gas Volumes given

Find the limiting reagent and the reactant in excess when 44.82L of CO(g) react completely with 11.205L of O₂(g) at S.T.P. (0^oC or 273K and 1atm or 101.3kPa)

Write the balanced chemical equation for the chemical reaction2CO(g) + O₂(g) -----> 2CO₂(g)

Calculate the available moles of each reactant in the chemical reaction

moles of CO = V ÷ 22.41
At S.T.P. 1 mole of gas
has a volume of 22.41Lmoles of CO = 44.82 ÷ 22.41
= 2mol

moles of O₂ = V ÷ 22.41
At S.T.P. 1 mole of gas
has a volume of 22.41Lmoles O₂ = 11.205 ÷ 22.41
= 0.5mol

Use the balanced chemical equation to determine the mole ratio of the reactants in the chemical reaction

CO : O₂ Or O₂ : CO

1 : ½ 1 : 2
Compare the available moles of each reactant to the moles required for complete reaction using the mole ratioIf all of the 2 moles of CO were to be used in the reaction it would require
½ x 2 = 1 mole of O₂ for the reaction to go to completion.
There are 0.5 moles of O₂ available which is less than the required 1 mole.
If all of the 0.5 moles of O₂ were to be used in the reaction it would require
2 x 0.5 = 1 mole of CO.
There are 2 moles of CO available which is more than the required 1 mole.
The limiting reagent is the reactant that will be completely used up during the chemical reaction. There will be some moles of the reactant in excess left over after the reaction has gone to completion.The limiting reagent is O₂,
all of the 0.5 moles of O₂ will be used up when this reaction goes to completion.
The reactant in excess is CO,
when the reaction has gone to completion there will be
2 - 1 = 1 mole of CO left over.

Chemical calculation based on mass-mass relationship

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Chemical calculation based on mass-mass relationship

Example:

Calculate the mass of lead chloride formed, by treating an aqueous solution of 6.62g of lead nitrate, with excess of hydrochloric acid.

(Relative atomic mass of: Pb=207, Cl=35.5, H=1, O=16, N=14)

Solution:

According to the equation,

207 + (14 +16 x 3) x 2 207 + (35.5 x 2)

207 + (14 + 48) x 2 207 + 71

207 + 62 x 2 278

207 + 124

331 g 278 g

331 g of lead nitrate yield 278 g of lead chloride. Hence, mass of lead chloride formed from 6.62 g of lead nitrate is -

Pb(NO₃)₂ : PbCl₂

331 : 278

6.62 : x

Mass of lead chloride formed = 5.56 g

Example:

A mixture of sodium chloride and anhydrous sodium carbonate has a mass of 5 g. It is dissolved in water, and treated with barium chloride solution. The mass of barium carbonate precipitated is 5.91 g. Calculate the mass of sodium chloride and the percentage of sodium carbonate present in the mixture.

(Relative atomic mass of C=12, O=16, Na=23, Cl=35.5, Ba=137)

Solution:

According to the equation,

(23 x 2) + {12 + (16 x 3)} 137 + {12 + (16 x 3)}

46 + 12 + 48 137 + 12 + 48

106 g 197 g

Mass of Na₂CO₃ needed to precipitate 197 g of BaCO₃ = 106 g.

Hence, mass of sodium carbonate needed to precipitate 5.91 g of barium carbonate=?

BaCO₃ : Na₂CO₃

197 : 106 g

5.91 g : x

Mass of sodium carbonate present in the mixture = 3.18 g

% of sodium carbonate present in the mixture

Mass of sodium chloride present in the mixture = 5 - 3.18 = 1.82 g

Example:

Calculate the mass of lead monoxide formed by the complete thermal decomposition of 3.425 g of red lead (Relative atomic mass of O=16, Pb=207).

Solution:

As per the equation,

(207 x 6) + (16 x 8) (207 x 6) + (16 x 6)

(1242 + 128) 1242 + 96

1370 1338

1370 g of red lead yield 1338 g of lead monoxide.

Mass of lead monoxide formed from 3.425 g of red lead = ?

Pb₃O₄ : PbO

1370 : 1338

3.425 : x

Mass of lead monoxide formed = 3.345 g

Example:

Calculate the percentage loss of mass suffered by potassium nitrate, when it is completely decomposed by heat. (Relative atomic mass of K=39, N=4, O=16).

Solution:

As per the equation,

2 {39 + 14 + (16 x 3)} (16 x 2)

2 (53 + 48) 32

2 (101) 32

202 32

Mass of oxygen liberated when 202 g of KNO₃ is heated = 32 g

Percentage loss of mass = 15.84%

Example:

Calculate the mass of copper nitrate obtained by treating 3.2 g of copper with excess concentrated nitric acid (Relative atomic mass of Cu=64, N=4, O=16).

Solution:

As per the equation,

64 64 + {14 + (16 x 3)} x 2

64 64 + (62 x 2 )

64 64 + 124

64 188

64 g of copper gives 188 g of copper nitrate.

Hence, mass of copper nitrate formed from 3.2 g of copper = ?

Cu : Cu(NO₃)₂

64 g : 188 g

3.2 g : x

Mass of copper nitrate formed = 9.4 g

Chemical calculation based on volume – volume relationship

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Chemical calculation based on volume – volume relationship

Example:

Calculate the approximate volume of air needed at STP, to oxidize 4480 litres of ammonia to nitric oxide in the Ostwald's process.

Solution:

As per the equation,

4 vols. : 5 vols.

4 x 22.4 litres : 5 x 22.4 litres

89.6 litres : 112 litres

Volume of oxygen required to oxidize 89.6 litres of ammonia = 122 litres

Volume of oxygen required to oxidize 4480 litres of ammonia =?

NH₃ : O₂

89.6 : 112

4480 : x

But oxygen forms only ^t^h the volume of air.

Hence volume of air needed = 5600 x 5 = 28000 litres

Volume of air = 28000 litres.

Example:

About 560 ml of carbon monoxide is mixed with 500 ml of oxygen and ignited. Calculate the volume of carbon dioxide formed, and the volume of oxygen left behind in excess.

Solution:

As per the equation,

2 vols. : 1 vol. 2 vol.

2 x 22400 ml : 22400 ml 2 x 22400 ml

44800 ml : 22400 ml 44800 ml

Volume of CO₂ formed by burning 2 volumes of CO = 2 volumes

Therefore, by igniting 560 ml of CO the volume of CO₂ formed = 560 ml

Volume of O₂ needed to burn 2 volumes of CO = 1 vol.

Volume of O₂ needed to burn 560 ml of CO = ?

2CO : O₂

2 vol. : 1 vol

560 ml : x

Hence volume of excess oxygen = 500 - 280 = 220 ml

Volume of carbon dioxide formed = 560 ml

Volume of excess oxygen = 220 ml.

Example:

Calculate the volume of gaseous products obtained by burning 4.8 litres of methane in requisite quantity of oxygen.

Solution:

As per the equation,

1 vol. 1 vol. + 2 vols.

1 vol. 3 vols.

Volume of gaseous products obtained by burning 1 volume of CH₄ = 3vols.

Volume of gaseous products obtained by burning 4.8 litres of CH₄ =?

Methane : (carbon dioxide + water)

1 vol. : 3 vols.

4.8 : x

Volumes of gaseous products formed = 14.4 litres.

Example:

Chlorine combines with excess ammonia to form ammonium chloride and nitrogen. Calculate the minimum volume of ammonia required to combine completely with 21 litres of chlorine.

Solution:

As per the equation,

8 vols. : 3 vols.

Volume of NH₃ required to combine with 3 volumes of Cl₂ = 8 vols

Volume of NH₃ required to combine with 21 litres of Cl₂ =?

NH₃ : Cl₂

8 : 3

x : 21

Minimum volume of ammonia required = 56 litres.

Chemical calculation based on mass volume relationship

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Chemical calculation based on mass volume relationship

Example:

Calculate the volume of carbon dioxide formed at STP in 'ml' by the complete thermal decomposition of 3.125 g of pure calcium carbonate (Relative atomic mass of Ca=40, C=12, O=16)

Solution:

As per the equation,

40 + 12 + (16 x 3) 1 mole

100g 22.4 litres

100 g 24400 ml

Volume of carbon dioxide formed from 100 g of CaCO₃ = 22400 ml

Volume of carbon dioxide from 3.125 g of CaCO₃ = ?

CaCO₃ : CO₂

100 g : 22400 ml

3.125 g : x

Volume of carbon dioxide formed = 700 ml.

Example:

Calculate the volume of ammonia formed at STP in 'ml' by treating 2.675 g of ammonium chloride with excess of calcium hydroxide (Relative Atomic Mass of N=14, O=16, H=1, Cl=35.5, Ca=40).

Solution:

As per the equation,

2 (14 + 4 + 35.5) 2 vols.(of NH₃)

107 2 x 22.4 litres

107 g 44.8 litres or 44800 ml

NH₄Cl : NH₃

107 g : 44800 ml

2.675 g : x

Volume of ammonia formed at STP = 1120 ml.

Example: 31Calculate the volume of sulphur dioxide formed at STP in 'ml', by treating 4.8 g of copper with excess of hot concentrated sulphuric acid. (Relative atomic mass of Cu=64, H=1, S=32, O=16).

Solution:

As per the equation,

64 g 1 volume (of SO₂)

64 g 22400 ml

Volume of SO₂ formed using 64 g of copper = 22400 ml

Volume of SO₂ formed using 4.8 g of copper = ?

Cu : SO₂

64 g : 22400 ml

4.8 g : x

Volume of sulphur dioxide formed at STP = 1680 ml.

Example:

Calculate the volume of water gas obtained by passing steam over 96 kg of white hot coke (Relative atomic mass of C=12, H=1, O=16).

Solution:

As per the equation,

12 g 1 vol. + 1 vol.

12 g (22.4 + 22.4) litres

12 g 44.8 litres

Volume of water gas formed from 12 g of coke = 44.8 litres

Volume of water gas formed from 96 kg of coke =?

C : Water gas

12 g : 44.8 litres

96000 g : x litres

Volume of water gas formed = 358400 litres.

Deduction of relationship between molecular mass and vapour density

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Deduction of relationship between molecular mass and vapour density

Relative Molecular Mass

Relative molecular mass is the ratio of the mass of one molecule of a substance to the mass ^t^h of a carbon atom, or 1 amu.

Vapour Density

Vapour density is the ratio of the mass of a volume of a gas, to the mass of an equal volume of hydrogen, measured under the same conditions of temperature and pressure.

According to Avogadro's law, equal volumes of all gases contain equal number of molecules.

Thus, let the number of molecules in one volume = n,

Cancelling 'n' which is common, we get

However, since hydrogen is diatomic.

When we compare the formulae of vapour density with relative molecular molecular mass, they can be represented as

We can therefore substitute the above equation, and arrive at the below formula

Now on cross multiplication, we have

2 x vapour density = Relative Molecular Mass of a gas

or,

Relative Molecular Mass = 2 x Vapour density

Deduction of molar volume of gases

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Deduction of molar volume of gases

Molar volume is the volume occupied by one mole of any gas at a definite pressure and temperature. It is denoted by V_m. Molar volume of the substance depends on temperature and pressure. The unit of molar volume is litre per mol or millilitre per mol.

As per Avogadro's law, equal volumes of all gases contain equal number of molecules, at a constant temperature and pressure. Therefore, equal number of molecules of any gas, must occupy the same volume, at constant temperature and pressure.

Standard Molar Volume

Standard molar volume of a gas is the volume occupied by 1 mole of any gas at 273 K and 1 atm pressure (STP). It is equal to 22.4 litres of 22,400 ml. It is the same for all gases.

Remember

S.T.P. = Standard Temperature and Pressure

Standard Temperature = 0^oC or 273 K

Standard Pressure = 1 atm or 760 mm of mercury

Calculation of Molar Volume

Example of oxygen

Mass of 1 litre of oxygen at STP = 1.429 g

Mass of 1 mol of oxygen = 32 g

Volume : Mass

1 litre : 1. 429 g

x : 32 g

The following table gives the relation between the Gram Molecular Weight (GMW), Number of moles, Molar Volume and the Number of particles for gases at STP.

Relationship between Various Parameters of Gases, at STP

Gas	Molecular Formula	GMW (in g)	No.Of Moles	Molar Volume dm³ or l	No.of moles in 1 mole
Hydrogen	H₂	2	1	22.4	6.023x 10²³
Oxygen	O₂	32	1	22.4	6.023x 10²³
Nitrogen	N₂	28	1	22.4	6.023x 10²³
Chlorine	Cl₂	71	1	22.4	6.023x 10²³
Carbon dioxide	CO₂	44	1	22.4	6.023x 10²³
Nitrogen dioxide	NO₂	46	1	22.4	6.023x 10²³
Ammonia	NH₃	17	1	22.4	6.023x 10²³
Methane	CH₄	16	1	22.4	6.023x 10²³
Sulphur dioxide	SO₂	64	1	22.4	6.023x 10²³

Problems Based on Molar Volume

Example:

Calculate the volume occupied by 3.4 g of ammonia at STP. (N=14, H=1)

Solution

Gram molecular mass of ammonia (NH₃) = (1 x 14) + (3 + 1) = 14 + 3 = 17 g

Mass of 1 mol of ammonia = 17g

Molar volume = 22.4 litres

Volume of 3.4 g of ammonia at STP = ?

Mass : Volume

17 g : 22.4 litre

3.4 g : x

Volume occupied by 3.4 g of ammonia at STP =4.48 litres

Example:

56 ml of carbon dioxide has a mass at 0.11 g at STP. What is the molar mass of the carbon dioxide?

Solution:

Mass of 56 ml of carbon dioxide at STP = 0.11 g

Mass of 22400 ml of carbon dioxide = ?

Mass : Volume

0.11 g : 56 ml

x g : 22400 ml

Mass of 22400 ml of carbon dioxide at STP = 44 g

Molar Mass of carbon dioxide = 44 g/mole.

Example:

One gram of pure sulphur dioxide has a volume of 350 ml at STP. What is the Relative Molecular Mass of sulphur dioxide?

Solution:

Volume of sulphur dioxide gas = 350 ml at STP

Mass of sulphur dioxide gas = 1 g

Mass of one mole of sulphur dioxide = x g/mole

Volume of 1 mole of sulphur dioxide = 22400 ml at STP

Mass : Volume

1 : 350 ml

x : 22400 ml

Mass of 1 mole of SO₂ = 64 g/mole

\ Relative molecular mass of sulphur dioxide = 64.

Example:

100 ml of carbon monoxide, has a mass of 0.125 g at STP. Calculate the mass of 1 mole of carbon monoxide.

Solution:

Volume of carbon monoxide = 100 ml

Mass of carbon monoxide = 0.125 g

Molar volume = 22400 ml

Mass of 1 mole of carbon monoxide =?

Volume : Mass

100 ml : 0.125 g

22400 ml : x g

Mass of one mole of carbon monoxide = 28 g/mole

\ Relative molecular mass of carbon monoxide = 28.

Concept of equivalent mass

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It is defined as the mass of an element/compound/ion which combines or displaces 1 part of hydrogen or 8 parts of oxygen or 35.5 parts of chlorine by mass.It is not always possible to apply this classic definition to determine equivalent weights of chemical entities. It is so because, we can not conceive of reactions involving chemical entities with three named reference of hydrogen, oxygen and chlorine. Generally, we are limited to determination of equivalent weights of elements and few compounds by using this definition of equivalent weight. A more workable definition is given as :

Equivalent weight,E=

Molecular weight

Valence factor

M_O

Clearly, determination of equivalent weight amounts to determining valence factor “x”. Here, we shall classify chemical entities and the techniques to determine “x”.

or,

Equivalent mass is equal to the molecular or atomic mass divided by the number of electrons involved in the reaction per molecule, atom or ion. For example in the reaction,

two electrons are needed to produce one molecule of hydrogen gas. So, 2 Faraday of electricity is needed to produce one mole of hydrogen gas.

Hence,

For the reaction,

per mole of copper atoms, 2 faradays are needed. So, the equivalent mass of copper is half of its atomic mass.

The equivalent mass of any species is not simply a property of the species, but depends upon the reaction in which it participates, i.e., one chemical species can have more than one value for its equivalent mass depending upon the reaction it participates.

The equivalent mass of a substance is the quantity of material deposited or dissolved by 1 F (= 96500 C) of electricity.

Equivalent weight of elements, and compounds

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Equivalent weight of elements, and compounds

Equivalent weight of an element

In the case of an element, the equivalent weight is defined as :

Equivalent weight,E=

Atomic weight

Valency

Note that atomic weight substitutes molecular weight and valency substitutes valence factor in the definition. Valencies of hydrogen, calcium and oxygen are 1,2 and 2 respectively. Hence, their equivalent weights are 1/1 =1, 40/2 = 20 and 16/2 = 8 respectively.

Equivalent weight of an acid

The valence factor of an acid is equal to its basicity. The basicity of an acid is equal to furnishable hydrogen ion (proton) in its aqueous solution. Importantly, basicity is not same as the number of hydrogen atoms in acid molecule. Consider acetic acid (CH3COOH). It contains 4 hydrogen atoms in it, but only 1 furnishable hydrogen ion. As such, basicity of acetic acid is 1. With this background, we define equivalent weight of an acid as :

Equivalent weight,E=

Molecular weight of acid

Basicity

Basicity of sulphuric acid is 2. Hence, equivalent weight of sulphuric acid (H₂SO₄) is (2X1 + 32 + 4X16)/2 = 98/2 = 49. Similarly, basicity of oxalic acid is 2. Hence, equivalent weight of oxalic acid (H₂C₂O₄ ) is (2X1 + 2X12 + 4X16)/2 = 90/2=45.

Phosphorous based acids like phosphoric acid (H₃PO₄ ), phosphorous acid (H₃PO₃) and hypo-phosphorous acid (H₃PO₂) need special mention here to understand their basicity. The structures of three acids are shown here. From the structure, it appears that these compounds may furnish OH ions, but bond strengths between phosphorous and oxygen (P-O) and phosphorous and hydrogen (P-H) are stronger than between oxygen and hydrogen (O-H) in –OH group. As such, these molecules release hydrogen ions from –OH group and behave as acid. Clearly, basicities of phosphoric acid (H₃PO₄), phosphorous acid (H₃PO₃) and hypo-phosphorous acid (H₃PO₂) are 3, 2 and 1 respectively.

Phosphorous based acids

Figure 1: Furnishable hydrogen ions of acids

Equivalent weight of a base

The valence factor of a base is equal to its acidity. The acidity of a base is equal to furnishable hydroxyl ion (OH-) in its aqueous solution. With this background, we define equivalent weight of a base as :

Equivalent weight,E=

Molecular weight of base

Acidity

Acidity of KOH is 1, whereas acidity of Ca(OH)₂ is 2. Hence, equivalent weight of KOH is (39 + 16 + 1)/1 = 56/1 = 56. Similarly, equivalent weight of Ca(OH)₂ is {40 + 2X(16+1)}/2 = 74/2=37.

Equivalent weight of a compound

The valence factor of a compound depends on the manner a compound is involved in a reaction. The compounds of alkali metal salts and alkaline earth metal salts are, however, constant. These compounds are ionic and they dissociate in ionic components in aqueous solution. In this case, valence factor is equal to numbers of electronic charge on either cation or anion.

Equivalent weight,E=

Molecular weight of compound

Numbers of electronic charge on cation or anion

The numbers of electronic charge on cation of NaHCO₃ is 1. Hence, equivalent weight of NaHCO₃ is (23 + 1 + 12 + 3X16)/1 = 84.

If we look at the defining ratio of equivalent weight of a compound (AB) formed of two radicals (say A and B), then we can rearrange the ratio as :

Equivalent weight, E=

Molecular weight of Radical A

Numbers of electronic charge

Molecular weight of Radical B

Numbers of electronic charge

Thus,

⇒Equivalent weight of AB=Equivalent weight of A+Equivalent weight of B

Equivalent weight of an ion

The valence factor of an ion is equal to numbers of electronic charge on the ion. Therefore, we define equivalent weight of an ion as :

Equivalent weight,E=

Molecular weight of ion

Numbers of electronic charge

The numbers of electronic charge on carbonate ion (CO₃²⁻ ) is 2. Hence, equivalent weight of carbonate ion is (12 + 3X16)/1 = 60/2 = 30. Similarly, equivalent weight of aluminum ion (Al³⁺) is 27/3 = 9.

Equivalent weight of an oxidizing or reducing agent

In a redox reaction, one of the reacting entities is oxidizing agent (OA). The other entity is reducing agent (RA). The oxidizer is recipient of electrons, whereas reducer is releaser of electrons. The valence factor for either an oxidizing or reducing agent is equal to the numbers of electrons transferred from one entity to another.

Equivalent weight,E=

Molecular weight of compound

Numbers of electrons transferred in redox reaction

Alternatively,

Equivalent weight,E=

Molecular weight of compound

Change in oxidation number in redox reaction

Potassium dichromate in acidic medium is a strong oxidizer. It means it gains electrons during redox reaction. Potassium dichromate in acidic solution results in :

K₂Cr₂O₇+14H⁺+6e⁻→2K⁺+2Cr³⁺+7H₂O

Equivalent weight ofK₂Cr₂O₇=

294.2

=49

Study of redox reaction is in itself an exclusive and extensive topic. We shall, therefore, discuss redox reaction separately.

Equivalent weight of elements, compounds ions Acid and Base

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Equivalent weight of elements, and compounds

Equivalent weight of an element

In the case of an element, the equivalent weight is defined as :

Equivalent weight,E=

Atomic weight

Valency

=

A

x

Note that atomic weight substitutes molecular weight and valency substitutes valence factor in the definition. Valencies of hydrogen, calcium and oxygen are 1,2 and 2 respectively. Hence, their equivalent weights are 1/1 =1, 40/2 = 20 and 16/2 = 8 respectively.

Equivalent weight of an acid

The valence factor of an acid is equal to its basicity. The basicity of an acid is equal to furnishable hydrogen ion (proton) in its aqueous solution. Importantly, basicity is not same as the number of hydrogen atoms in acid molecule. Consider acetic acid (CH3COOH). It contains 4 hydrogen atoms in it, but only 1 furnishable hydrogen ion. As such, basicity of acetic acid is 1. With this background, we define equivalent weight of an acid as :

Equivalent weight,E=

Molecular weight of acid

Basicity

Basicity of sulphuric acid is 2. Hence, equivalent weight of sulphuric acid (H₂SO₄) is (2X1 + 32 + 4X16)/2 = 98/2 = 49. Similarly, basicity of oxalic acid is 2. Hence, equivalent weight of oxalic acid (H₂C₂O₄ ) is (2X1 + 2X12 + 4X16)/2 = 90/2=45.

Phosphorous based acids like phosphoric acid (H₃PO₄ ), phosphorous acid (H₃PO₃) and hypo-phosphorous acid (H₃PO₂) need special mention here to understand their basicity. The structures of three acids are shown here. From the structure, it appears that these compounds may furnish OH ions, but bond strengths between phosphorous and oxygen (P-O) and phosphorous and hydrogen (P-H) are stronger than between oxygen and hydrogen (O-H) in –OH group. As such, these molecules release hydrogen ions from –OH group and behave as acid. Clearly, basicities of phosphoric acid (H₃PO₄), phosphorous acid (H₃PO₃) and hypo-phosphorous acid (H₃PO₂) are 3, 2 and 1 respectively.

Figure 1: Furnishable hydrogen ions of acids

Phosphorous based acids

Equivalent weight of a base

The valence factor of a base is equal to its acidity. The acidity of a base is equal to furnishable hydroxyl ion (OH-) in its aqueous solution. With this background, we define equivalent weight of a base as :

Equivalent weight,E=

Molecular weight of base

Acidity

Acidity of KOH is 1, whereas acidity of Ca(OH)₂ is 2. Hence, equivalent weight of KOH is (39 + 16 + 1)/1 = 56/1 = 56. Similarly, equivalent weight of Ca(OH)₂ is {40 + 2X(16+1)}/2 = 74/2=37.

Equivalent weight of a compound

The valence factor of a compound depends on the manner a compound is involved in a reaction. The compounds of alkali metal salts and alkaline earth metal salts are, however, constant. These compounds are ionic and they dissociate in ionic components in aqueous solution. In this case, valence factor is equal to numbers of electronic charge on either cation or anion.

Equivalent weight,E=

Molecular weight of compound

Numbers of electronic charge on cation or anion

The numbers of electronic charge on cation of NaHCO₃ is 1. Hence, equivalent weight of NaHCO₃ is (23 + 1 + 12 + 3X16)/1 = 84.

If we look at the defining ratio of equivalent weight of a compound (AB) formed of two radicals (say A and B), then we can rearrange the ratio as :

Equivalent weight, E=

Molecular weight of Radical A

Numbers of electronic charge

+

Molecular weight of Radical B

Numbers of electronic charge

Thus,

⇒Equivalent weight of AB=Equivalent weight of A+Equivalent weight of B

Equivalent weight of an ion

The valence factor of an ion is equal to numbers of electronic charge on the ion. Therefore, we define equivalent weight of an ion as :

Equivalent weight,E=

Molecular weight of ion

Numbers of electronic charge

The numbers of electronic charge on carbonate ion (CO₃²⁻ ) is 2. Hence, equivalent weight of carbonate ion is (12 + 3X16)/1 = 60/2 = 30. Similarly, equivalent weight of aluminum ion (Al³⁺) is 27/3 = 9.

Equivalent weight of an oxidizing or reducing agent

In a redox reaction, one of the reacting entities is oxidizing agent (OA). The other entity is reducing agent (RA). The oxidizer is recipient of electrons, whereas reducer is releaser of electrons. The valence factor for either an oxidizing or reducing agent is equal to the numbers of electrons transferred from one entity to another.

Equivalent weight,E=

Molecular weight of compound

Numbers of electrons transferred in redox reaction

Alternatively,

Equivalent weight,E=

Molecular weight of compound

Change in oxidation number in redox reaction

Potassium dichromate in acidic medium is a strong oxidizer. It means it gains electrons during redox reaction. Potassium dichromate in acidic solution results in :

K₂Cr₂O₇+14H⁺+6e⁻→2K⁺+2Cr³⁺+7H₂O

Equivalent weight ofK₂Cr₂O₇=

294.2

6

=49

Study of redox reaction is in itself an exclusive and extensive topic. We shall, therefore, discuss redox reaction separately.

EXERCISE 1 – trends in Period 3

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1.4 EXERCISE 1 – trends in Period 3

1. State and explain the following trends:

a) atomic size across Period 3

b) first ionisation energy across Period 3

c) electronegativity across Period 3

d) melting point from sodium to aluminium

e) electrical conductivity from sodium to aluminium

2. Explain the following:

a) the melting point of silicon is the highest in Period 3

b) the melting point of phosphorus is lower than silicon

c) the melting point of phosphorus is lower than that of sulphur

d) the boiling point of chlorine is greater than that of argon, but lower than that of sulphur

e) neither silicon nor sulphur conduct electricity

Answers to 1.4 exercises

1.4 Exercise 1

1. a) decreases

proton number increases

shielding stays the same

so attraction to nucleus increases

b) increases

proton number increases

shielding stays the same

so attraction to nucleus increases

so more energy required to remove electron

c) increases

proton number increases

shielding stays the same

so ability to attract electrons in a covalent bond increases

d) increases

size of ions decreases

and charge on ions increases

so attraction between ions and delocalised electrons increases

so more energy is required to separate them

e) increases

number of delocalised electrons per atom increases

so more electrons are free to move

2. a) Si has a giant covalent structure

every Si atom attached to 4 others by covalent bonds

the covalent bonds are strong

so a lot of energy is required to break them

b) P is simple molecular

Van der Waal’s forces between P molecules are weaker

Than covalent bonds between Si atoms

So much less energy is required to break them

c) Phosphorus contains P molecules

which have a smaller surface area and less electrons

than the S molecules in sulphur

so the Van der Waal’s forces are weaker in phosphorus

and less energy is required to overcome them

d) Chlorine contains Cl molecules

which have a smaller surface area and less electrons

than the S molecules in sulphur

but a larger surface area and more electrons

than the Ar atoms in argon

so the energy required to separate Cl molecules is greater

than the energy needed to separated Ar atoms but less

than the energy needed to separated S molecules

e) neither silicon or sulphur contain ions or free electrons

A Few Practice Questions about Periodicity! NOTE: “more stable” = lower energy state.

“less stable” = higher energy state

1. The most important factor in determining the variation in size of atoms as you move from left to right within a series is the:

a. increase in nuclear charge b. increase in number of valence electrons

c. increase in the radii of electron shells d. increased shielding effect

e. decrease in nuclear charge

2. The major contributing factor to the decrease in ionization energy as you move to the bottom of a family is :

a. nuclear charge b. valence electrons

c. more inner shell electrons d. reduced shielding effect

3. As the atomic numbers of the elements in Group 16 (oxygen family)increase, the shielding effect:

a. decreases b. increases c. remains the same d. breaks

4. Ions formed by the gain of an electron are always more _____ than the atom they came from.

a. stable b. unstable

c. electronegative d.I can't tell because I don't know what the atom is.

5. The energy value for electron affinity within a family generally becomes ______ with increasing atomic number.

a. more negative b. less negative c. constant

6. Based on valence electrons, what is the most probable oxidation state of N?

a. -3 b. -2 c. +3 d. 0

7. You do an experiment which involves adding an electron to an atom of He. You make the following assumption in regard to its stability:

a. the ion is more stable b. the ion is less stable c. the stability doesn't change

d. You can't tell what happened because this statement doesn't give enough information.

8. Which of the following elements is most metallic? a. He b. F c. S d. Te e. Sb

9. Which of the following has a larger radius?

a. chlorine atom b. chlorine anion c. chlorine cation d. they're the same

10. Which of the following is least likely to gain electrons in a chemical reaction?

a. potassium b. aluminum c. fluorine d. oxygen

11. High ionization energy is characteristic of:

a. metals b. non-metals c. metalloids d. liquids

12. Which of the following trends generally have an opposite pattern from atomic radius?

a. electronegativity b. electron affinity c. ionization energy d. all of these

e. none of these

13. If energy is released during a process, the resulting particle is:

a. more stable b. less stable c. about the same d. irrelevant

14. An atom which lost energy during an electron transfer:

a. gained an electron b. lost an electron c. lost a proton

d. there is insufficient information to answer this question.

15. Which of the following elements would you expect to have a positive electron affinity?

a. K b. Ne c. O d. F

16. Compared to the stability of the original atom, the stability of its ion formed by the gain of an electron is:

a. always less b. sometimes less c. the same d. always greater

17. Which of the following has the largest radius? a. Ne b. F^- c. Na⁺ d. Mg⁺⁺

18. Which element would form an ion which is isoelectronic with Ne and would bond with oxygen in a 1:2 ratio:

a. F b. Na c. Mg d. Al e. K

19. The greatest increase in ionization energy for the element B comes with the removal of the _________ electron?

a. 2^nd b. 3^rd c. 4^th d. 5^th e. 6^th

20. An atom of Neon which gained energy during an electron transfer:

a. gained an electron b. lost an electron

c. lost a proton d. there is insufficient information to answer this question.

21. What element will form an ion which is isoelectronic with Ne and will combine with phosphorous in a 3 : 2 ratio? (2 atoms of phosphorous)

Answers: 1a 2c 3b 4d 5b 6a 7b 8e 9b 10a 11b 12d 13a 14a 15b 16b 17b 18b 19c 20d 21Mg

Isoelectronic Species

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Isoelectronic Species

For each of the following ions:

Tell which noble gas it is isoelectronic with

Provide names

Provide electron configurations and orbital diagrams

a. Br^-

b. Mg²⁺

c. N^3-

d. Ba²⁺

e. P^3-

f. Li⁺

g. O^2-

h. K⁺

i. Sr²⁺

j. S^2-

Answers are below. (Orbital notations are coming)

a. Br^-is isoelectronic with krypton.

Bromide

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶

b. Mg²⁺ is isoelectronic with neon.

Magnesium ion

1s² 2s² 2p⁶

c. N^3- is isoelectronic with neon.

Nitride

1s² 2s² 2p⁶

d. Ba²⁺ is isoelectronic with krypton.

Barium ion

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶

e. P^3- is isoelectronic with argon.

Phosphide

1s² 2s² 2p⁶ 3s² 3p⁶

f. Li+ is isoelectronic with helium.

Lithium ion

1s²

g. O^2- is isoelectronic with neon.

Oxide

1s² 2s² 2p⁶

h. K⁺ is isoelectronic with argon.

Potassium ion

1s² 2s² 2p⁶ 3s² 3p⁶

i. Sr²⁺ is isoelectronic with krypton.

Strontium ion

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶

j. S^2- is isoelectronic with argon.

Sulfide

1s² 2s² 2p⁶ 3s² 3p⁶

intermolecular forces

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A. List the types of intermolecular forces that exist between the molecules:

(a) Benzene (C6H6) molecules are nonpolar. Only dispersion forces will be present.

(b) Chloroform (CH3Cl) molecules are polar (why?). Dispersion and dipole-dipole forces will be present.

(c) Phosphorus trifluoride (PF3) molecules are polar. Dispersion and dipole-dipole forces will be present.

(d) Sodium chloride (NaCl) is an ionic compound. Ion-ion and dispersion forces will be present.

(e) Carbon disulfide (CS2) molecules are nonpolar. Only dispersion forces will be present.

1: PCl5 : Dispersion

2. CS2: Dispersion

3: CH3CH2OH: Dispersion, Dipole-Dipole, Hydrogen

4 SF4: Dispersion, Dipole-Dipole

B. Which has the strongest intermolecular forces of attraction?

1. CH3CH2OH or CH3CH2Br

I chose:CH3CH2OH

2. SI4 or CI4

I chose Sulfur Tetraiodide

C. Which has the lowest boiling point?

1. CH3CH2Cl or CH3CH2OH

I chose: CH3CH2Cl

2. NH3 or PH3

I chose: NH3

D. Which has the highest vapor pressure?

1. CH3Br or CH3Cl

I chose: CH3Cl

2. F2 or HCl

I chose: F2

Vector

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Vectors and Vector Arithmetic

Scalar quantities have a magnitude only: examples are mass, temperature, length, time, energy.
Vector quantities have a magnitude and a direction: examples are velocity ("50 m/s east"), force ("12 newtons up"), displacement ("18 inches to the left").
Vectors are identified either by boldface type, or by an arrow over the symbol.
One may also express vectors as a set of components:
```
            vector A   =  ( Ax , Ay )
```
Vector arithmetic is not as simple as scalar arithmetic: the results depend not only on the magnitudes of the inputs, but also on their directions.
To add or subtract vectors, one may use a graphical method. Place the vectors head-to-tail, and the difference between the tail of the first and the head of the last is the vector sum.

One may also add or subtract vectors by breaking each down into components, then adding or subtracting the component values:

            vector A   =  ( 12,  5 )
            vector B   =  ( -4, 20 )

                            x component        y component
                       --------------------------------------
              A + B          12 + (-4)            5 + 20
                                8                  25

        vector (A+B)   =  (  8, 25 )

One can use trigonometry to calculate the components of a vector, or to calculate the magnitude and direction of a vector from its components.

Viewgraphs