Fifteen Examples

Assume, unless otherwise told, that in all problems water is the solvent.

**Example #1:** Given a density of 1.836 g/mL and a mass percent of H_{2}SO_{4} of 96.00%, find the molarity, molality, and mole fraction. The molar mass of water is 18.015 g/mol and the molar mass of sulfuric acid is 98.078 g/mol. (Two different starting assumptions are shown.)

**Solution assuming a certain volume of solution is present:**

1) Assume that a volume of 1.000 L of the solution is present. Determine the total mass of the solution:

(1.836 g/mL) (1000. mL) = 1836 g

2) Determine the mass of each component of the solution:

H_{2}SO_{4}---> (1836 g) (0.9600) = 1762.56 g

H_{2}O ---> 1836 g minus 1762.56 g = 73.44 g

3) Determine the moles of each component of the solution:

H_{2}SO_{4}---> 1762.56 g / 98.078 g/mol = 17.9710 mol

H_{2}O ---> 73.44 g / 18.015 g/mol = 4.0766 mol

4) Determine the mole fraction of each component of the solution:

17.9710 mol + 4.0766 mol = 22.0476 molH

_{2}SO_{4}---> 17.9710 mol / 22.0476 mol = 0.8151

H_{2}O ---> 4.0766 mol / 22.0476 mol = 0.1849Often the last mole fraction is obtained by subtraction:

H

_{2}O ---> 1 − 0.8151 = 0.1849 <--- The 1 is NOT one significant figure.

5) Determine the molarity of the solution:

17.9710 mol / 1.000 L = 17.97 M (to four sig figs)

6) Determine the molality of the solution:

73.44 g = 0.07344 kg17.9710 mol / 0.07344 kg = 244.7 m

**Solution assuming a certain mass of the solution is present:**

1) Assume 100.0 g of the solution is present. Determine the mass of each solution component:

H_{2}SO_{4}---> (100.0 g) (0.9600) = 96.00 g

H_{2}O ---> (100.0 g) (0.0400) = 4.00 g

2) Determine the mole fraction:

H_{2}SO_{4}---> 96.00 g / 98.078 g/mol = 0.978813 mol

H_{2}O ---> 4.00 g / 18.015 g/mol = 0.222037 mol0.978813 mol + 0.222037 mol = 1.20085 mol

H

_{2}SO_{4}---> 0.978813 mol / 1.20085 mol = 0.8151

H_{2}O ---> 0.222037 mol / 1.20085 mol = 0.1849or, the H

_{2}O can be obtained by subtraction:1 − 0.8151 = 0.1849

3) Determine the molality:

4.00 g = 0.00400 kg0.978813 mol / 0.00400 kg = 244.7 m

4) Determine the molarity:

100.0 g / 1.836 g/mL = 54.46623 mL = 0.05446623 L0.978813 mol / 0.05446623 L = 17.97 M

5) By the way, you could consider this solution to be some water (the solute) dissolved in some sulfuric acid (the solvent). Calculate the molality of the water:

1762.56 g = 1.76256 kg4.0766 mol / 1.76256 kg = 2.313 m

**Example #2:** Given a density of 1.769 g/mL, and a H_{2}SO_{4} mole fraction of 0.5000, find the molality, molarity, and mass percent.

**Solution:**

1) We will use a mole fraction of 0.5000 to mean 0.5000 mole is present in a total of 1.0000 mole of solution. Determine the mass of each mole fraction:

H_{2}SO_{4}---> (0.5000 mol) (98.078 g/mol) = 49.039 g

H_{2}O ---> (0.5000 mol) (18.015 g/mol) = 9.0075 g

Comment: A mole fraction of 0.50 could mean 1.0 mol of one component in 2.0 total moles. Using 0.5 and 1 is the simplest meaning of a mole fraction of 0.5. We could have used any paring of numbers that gives a mole fraction of 0.5. The final answers would be the same, but the numbers in the calculations would be different.

2) Determine the mass percent of each component:

49.039 g + 9.0075 g = 58.0465 gH

_{2}SO_{4}---> 49.039 g / 58.0465 g = 84.48%

H_{2}O ---> 9.0075 g / 58.0465 g = 15.52%Often the last mass percent is obtained by subtraction:

H

_{2}O ---> 100.00 − 84.48 = 15.52

3) Determine the molality:

9.0075 g = 0.0090075 kg0.5000 mol / 0.0090075 kg = 55.51 m

4) Determine the molarity:

58.0465 g / 1.769 g/mL = 32.81 mL = 0.03281 L0.5000 mol / 0.03281 L = 15.24 M

**Example #3:** Given a density of 1.059 g/mL and a H_{2}SO_{4} molarity of 1.000 M, find the molality, mole fraction, and mass percent.

**Solution:**

1) Assume 1.0000 L of the solution is present. Determine the mass of the solution:

(1.059 g/mL) (1000.0 mL) = 1059 g

2) Determine the mass percent of each component:

H_{2}SO_{4}---> 98.078 g (remember, it's a 1 M solution)

H_{2}O ---> 1059 g − 98.078 g = 960.922 gH

_{2}SO_{4}---> 100 − 90.74 = 9.26%

H_{2}O ---> 960.922 g / 1059 g = 90.74%Note that I calculated the larger value by division and the smaller value by subtraction.

3) Determine the mole fraction:

H_{2}SO_{4}---> 98.078 g / 98.078 g/mol = 1.000 mol (or just remember it's a 1 M solution)

H_{2}O ---> 960.922 g / 18.015 g/mol = 53.349 mol53.349 mol + 1.000 mol = 54.349 mol

H

_{2}SO_{4}---> 1 − 0.9816 = 0.0184

H_{2}O ---> 53.349 mol / 54.349 mol = 0.9816

4) Determine the molality:

960.922 g = 0.960922 kg1.000 mol / 0.960922 kg = 1.041 m

**Example #4:** Given a density 1.122 g/mL and a H_{2}SO_{4} molality of 4.500 m, find the molarity, mole fraction and mass percent.

**Solution:**

1) The given molality means 4.500 mol dissolved in 1.000 kg of water. Determine the mass of each component:

H_{2}SO_{4}---> (4.500 mol) (98.078 g/mol) = 441.351 g

H_{2}O ---> 1.000 kg = 1000. g

2) Determine mass percentages:

1000. g + 441.351 g = 1441.351 g (total mass of the solution)H

_{2}O ---> 1000. g / 1441.351 g = 69.38%

H_{2}SO_{4}---> 100 − 69.38 = 30.62%

3) Determine mole fraction:

H_{2}SO_{4}---> 4.500 mol

H_{2}O ---> 1000. g / 18.015 g/mol = 55.509 mol55.509 mol + 4.500 mol = 60.009 mol

H

_{2}SO_{4}---> 1 − 0.9250 = 0.0750

H_{2}O ---> 55.509 mol / 60.009 mol = 0.9250

4) Determine the molarity:

1441.351 g / 1.122 g/mL = 1296.179 mL = 1.296179 L4.500 mol / 1.296179 L = 3.472 M

**Example #5:** Calculate the percent cadmium(II) bromide by mass in an aqueous solution with a solute mole fraction of 0.114 and a density of 1.047 g/mL.

**Solution:**

1) Determine mole fraction of the water:

1 − 0.114 = 0.886

2) Assume a solution with 0.114 mol of CdBr_{2} and 0.886 mol of water is present. Determine the mass of each solution component:

CdBr_{2}---> (0.114 mol) (272.218 g/mol) = 31.033 g

H_{2}O ---> (0.886 mol) (18.015 g/mol) = 15.889 g

3) The mass percent of CdBr_{2} is:

[31.033 g / (31.033 g + 15.889 g)] * 100 = 66.1%The solution density is not required.

Notice that the density is given in each of the above examples. This provides a necessary bridge between the volume-based concentration unit of molarity and the other concentration units (molality, mole fraction and mass percent) none of which uses volume in its definition.

If you were __only__ given two of these: (1) mass percent or (2) molality or (3) mole fraction (and no density), you could not get the molarity (or the density).

However, suppose you are given one of these: (1) mass percent or (2) molality or (3) mole fraction and then also given molarity instead of the density. Could you get to the density and the other units? In other words, can we swap density and molarity in the given part of the problem? Let's see . . . .

**Example #6:** Reagent grade nitric acid is 70.40% HNO_{3} (63.0119 g/mol) by mass and its molarity is 16.00 M. Calculate the density, molality and mole fraction of nitric acid in the solution.

**Solution:**

1) 16.0 molar means this:

16.0 moles of HNO_{3}

1.000 L (which equals 1000. mL) of solution.

2) The key point is that the 16.00 moles of HNO_{3} is 70.40% of the entire mass of the 1000. mL of solution.

(63.0119 g/mol) (16.0 mol) = 1008.19 g1008.19 g / 0.7040 = 1432 g (this is the total mass of the solution)

3) The density is:

1432 g / 1000. mL = 1.432 g/mL

4) Molality:

1432 g − 1008.19 g = 423.81 g = 0.42381 kg16.00 mol / 0.42381 kg = 37.75 m

5) Mole fraction

423.81 g / 18.015 g/mol = 23.5254 mol (of water)16.00 mol + 23.5254 mol = 39.5254 mol (total moles)

23.5254 mol / 39.5254 mol = 0.5952 (mole fraction of water)

1 − 0.5952 = 0.4048 (mole fraction of the nitric acid)

Here's the question again:

. . . can we swap density and molarity in the given part of the problem?

And the answer is a very firm YES. Let's do another . . . .

**Example #7:** Reagent grade nitric acid (HNO_{3}, MW = 63.0119 g/mol) has a molarity of 16.00 M and its molality is 37.75 m. Calculate the density, mass percent and mole fraction of nitric acid in the solution.

**Solution:**

1) Use the molality:

37.75 mol (of HNO_{3})

1.000 kg of solvent (water)

2) Compute the mass of the above solution:

(37.75 mol) (63.0119 g/mol) = 2378.7 g1000 g + 2378.7 g = 3378.7 g

3) Compute the mass percent (I'll do just the nitric acid):

2378.7 g / 3378.7 g = 70.40%

4) Compute the mole fraction of the nitric acid:

HNO_{3}---> 37.75 mol

H_{2}O ---> 1000 g / 18.015 g/mol = 55.509 mol55.509 mol + 37.75 mol = 93.259 mol

37.75 mol / 93.259 mol = 0.4048

5) Use the molarity to get the density of the solution:

37.75 mol / x = 16.00 mol/Lx = 2.359375 L

3378.7 g / 2359.375 mL = 1.432 g/mL

Turns out everything works just fine if the density and the molarity are the two values given. See Example #3 for another like #8.

**Example #8:** Reagent grade nitric acid (HNO_{3}, MW = 63.0119 g/mol) has a molarity of 16.00 M and a density of 1.432 g/mL. Calculate the molality, mass percent and mole fraction of nitric acid in the solution.

**Solution:**

1) Assume 1.000 L of the solution is present. Determine its mass:

(1.432 g/mL) (1000. mL) = 1432 g

2) Determine the mass percent (just the nitric acid):

(16.00 mol) (63.0119 g/mol) = 1008.19 g1008.19 g / 1432 g = 70.40%

3) Molality:

1432 − 1008.19 = 423.81 g = 0.42381 kg16.00 mol / 0.42381 kg = 37.75 m

4) Mole ratio (of just the water):

HNO_{3}---> 1008.19 g / 63.0119 g/mol = 16.00 mol

H_{2}O ---> 423.81 g / 18.015 g = 23.5254 mol23.5254 mol + 16.00 mol = 39.5254 mol

23.5254 / 39.5254 = 0.5952

**Example #9:** What is the molarity of a 30.0% (w/w) hydrogen peroxide solution?

Comment: note how the density has to be looked up (probably elsewhere in the textbook chapter this question came from) in order to solve the problem. (The ChemTeam did not write this question.) If you do not realize that you have to supply the density, you'd think it's an impossible question.

**Solution:**

1) Looking on the Internet, the density is found to be 1.11 g/mL. The reason the density is available is because 30% H_{2}O_{2} is commercially available. Its density is listed on the Material Safety Data Sheet that chemical supplies and purchasers are required to have. Here is one for 30% H_{2}O_{2}.

2) We will assume 1.00 L of the solution is present. (This is a convenient volume to take because you want molarity, which is defined as moles solute / liter solution.)

3) Compute the mass of 1.00 L of solution:

(1.11 g/mL) (1000 mL) = 1110 g

4) Compute the mass of H_{2}O_{2} in the liter of solution

(1110 g) (0.300) = 333 g H_{2}O_{2}

5) Compute moles of H_{2}O_{2} in the solution:

333 g / 34.0138 g/mol = 9.79 mol

6) Compute the molarity:

9.79 mol / 1.00 L = 9.79 M

**Example #10:** A 1.55 m solution of glucose (C_{6}H_{12}O_{6}) is present. Determine the mole ratio of each solution component as well as the mass percent.

**Solution:**

1) 1.55 m means 1.55 mole of glucose dissolved in 1.00 kg of water. Determine the mass percents:

glucose ---> (1.55 mole) (180.1548 g/mol) = 279.23994 g

H_{2}O ---> 1000 g1000 g + 279.23994 g = 1279.23994 g

glucose ---> 100 − 71.87 = 21.83%

H_{2}O ---> 1000 g / 1279.23994 g = 78.17%

2) Determine the mole ratios:

glucose ---> 1.55 mole

H_{2}O ---> 1000 g / 18.015 g/mol = 55.5091.55 + 55.509 = 57.059 mol

glucose ---> 1 − 0.9728 = 0.0272

H_{2}O ---> 55.509 / 57.059 = 0.9728

Comment: Give me the density and I can compute the molarity. Give me the molarity and I can compute the density. However, since neither one is present, the above problem is as far as we can go.

**Example #11:** Determine the mole fraction of H_{2}O and CH_{3}OH in a solution whose molality is 2.00 m.

**Solution:**

1) Let us consider water to be the solvent. This means:

2.00 moles of CH_{3}OH dissolved in 1.00 kg of H_{2}O

2) Determine moles of water:

1000 g ––––––––– = 55.509 mol 18.015 g/mol

3) Determine total moles present:

55.509 mol + 2.00 mol = 57.509 mol

4) Determine the mole fraction of water:

55.509 mol ––––––––– = 0.965 57.509 mol

5) The mole fraction of methyl alcohol can be arrived at by subtraction:

1 − 0.965 = 0.035You can also do it by division, but be aware that the mole fractions may not add up to 1, due to rounding errors.

**Examples #12 - 14:** Fill in the blanks in the table for aqueous solutions of the compounds shown. The density of the solution is in g/cm^{3}.

Compound MW density m %(w/w) χ M NH _{3}17.0307 0.9228 20.00 Na _{2}Cr_{2}O_{7}261.965 1.1260 0.7271 NH _{2}CONH_{2}60.0556 1.0276 1.882

**Solutions for ammonia:**

1) Assume 100.0 g of the solution is present. The 20.00%(w/w) tells us this:

ammonia: 20.00 g

water: 80.00 g

2) For the molality calculation, we need to know the moles of ammonia:

20.00 g ––––––––– = 1.17435 mol 17.0307 g/mol Then, calculate the molality:

1.17435 mol ––––––––– = 14.68 m 0.08000 kg

3) For the mole fractions, we need to know the moles of water, so we can then determine the total moles in the solution:

80.00 g ––––––––– = 4.440744 mol 18.015 g/mol total moles ---> 1.17435 mol + 4.440744 mol = 5.615094 mol

χ

_{water}---> 4.440744 mol / 5.615094 mol = 0.7908

χ_{ammonia}---> 1 − 0.7908 = 0.2092

4) For the molarity, we first use the density and 100.0 g of solution to get the volume of the solution:

100.0 g ––––––––– = 108.366 cm ^{3}0.9228 g/cm ^{3}Calculate the molarity:

1.17435 mol ––––––––– = 10.84 M 0.108366 L

**Solutions for sodium dichromate:**

1) The molality tells us this:

Na_{2}Cr_{2}O_{7}---> 0.7271 mol

H_{2}O ---> 1.000 kg

2) Change moles to grams:

(0.7271 mol) (261.965 g/mol) = 190.475 g

3) Calculate the mass percentages:

H_{2}O ---> 1000. g / 1190.475 g = 84.00%

Na_{2}Cr_{2}O_{7}---> 1 − 0.8400 = 16.00%

4) For the mole fraction, we know how many moles of Na_{2}Cr_{2}O_{7} are in the solution (0.7271 mol). Determine the moles of water:

1000. g ––––––––– = 55.5093 mol 18.015 g/mol mole fraction of water ---> 55.5093 mol / 56.2364 mol = 0.9871

mole fraction of the sod. dichromate ---> 1 - 0.9871 = 0.0129

5) For the molarity calculation, we first use total mass of the solution (1190.475 g; calculated for the mass percents above) to determine the volume of the solution:

1190.475 g ––––––––– = 1057.26 cm ^{3}= 1.05726 L1.1260 g/cm ^{3}Calculate the molarity:

0.7271 mol ––––––––– = 0.6877 M 1.05726 L

**Solutions for urea:**

1) Let's assume 1.000 L of solution is present. Using the density, let's determine the mass of the solution:

1000. cm ^{3}––––––––– = 973.1413 g 1.0276 g/cm ^{3}

2) In that 1.000 L of solution, there is 1.882 mole of urea. Determine its mass:

(1.882 mol) (60.0556 g/mol) = 113.0246 g

3) Determine the molality:

973.1413 g − 113.0246 g = 860.1167 g of water

1.882 mol ––––––––– = 2.188 m 0.8601167 kg

4) To determine the mole fractions, we need to know how many moles of water are present:

860.1167 g ––––––––– = 47.7445 mol 18.015 g/mol Calculate the mole fractions:

water ---> 47.7445 mol / 49.6265 mol = 0.9621

urea ---> 1 − 0.9621 = 0.0379

5) The weight percents:

water ---.> 860.1167 g / 973.1413 g = 88.38%

urea ---> 11.62%

**Example #15:** Dilute 38%(w/w) hydrochloric acid to a pH of 1.00 with water. The density of 38% HCl is 1.19 g/mL

**Solution:**

1) Let's start by analyzing what we know about the end result.

(a) We will assume the final volume to be 1.00 L.

(b) The pH of 1.00 means the final solution contains 0.10 mole of HClthe result in (b) comes from this:[H

^{+}] = 10^{-pH}= 10^{-1.00}= 0.10 M(c) Remember, HCl is a strong acid, so it ionizes 100%

(d) The molar mass of HCl is 36.461 g/mol, so 3.6461 g of HCl is required. (I'll round off to three sig figs at the end.)

2) Our problem now becomes: what mass of 38% HCl solution is required to deliver 3.6461 g of HCl? A simple ratio and proportion is all that is required:

38 g 3.6461 g ––––– = ––––––– 100 g x x = 9.595 g

3) We can easily weigh out 9.595 g of the 38% solution and dilute that amount to 1.00 L of solution. The 1.00 L of solution contains 3.6461 grams of HCl (0.10 mole of HCl). The molarity of the diluted solution is 0.10 M and its pH is this:

pH = −log [H^{+}] = −log 0.10 = 1.00

4) Holy Moly! I didn't use the density! What did I do wrong?

We would only require the density if the problem had asked for the volume of HCl solution required. I admit that weighing out a mass of solution is not the usual thing to do, but it is certainly possible and it is within the bounds of this particular problem (since the unit of the answer was not specified).

5) What if the problem had asked for the volume of 38% solution to be diluted? In this case, the density is required

9.595 g ––––––––– = 8.063 mL 1.19 g/mL To three sig figs, the answer would be 8.06 mL.