Calculations involving molality, molarity, density, mass percent, mole ratio
Problems #11 - 25

Ten Examples

Problems 1 - 10

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Problem #11: Calculate the molarity and mole fraction of acetone in a 2.28-molal solution of acetone (CH3COCH3) in ethanol (C2H5OH). (Density of acetone = 0.788 g/cm3; density of ethanol = 0.789 g/cm3.) Assume that the volumes of acetone and ethanol add.

Solution for molarity:

Remember, 2.28-molal means 2.28 moles of acetone in 1.00 kilogram of ethanol.

1) Determine volumes of acetone and ethanol, then total volume:

acetone

2.28 mol x 58.0794 g/mol = 132.421 g

132.421 g divided by 0.788 g/cm3 = 168.047 cm3

ethanol

1000 g divided by 0.789 g/cm3 = 1267.427 cm3

total volume

168.047 + 1267.427 = 1435.474 cm3

2) Determine molarity:

2.28 mol / 1.435 L = 1.59 M

Solution for mole fraction:

1) Determine moles of ethanol:

1000 g / 46.0684 g/mol = 21.71 mol

2) Determine mole fraction of acetone:

2.28 / (2.28 + 21.71) = 0.0950

Problem #12: Calculate the normality of a 4.0 molal sulfuric acid solution with a density of 1.2 g/mL.

Reminders:

N = #equivalents / L solution
#equivalents = molecular weight / n (n = number of H+ or OH¯ released per dissociation.)
molal = moles solute / kg solvent

Solution:

1) Determine grams of H2SO4 present:

4.0 molal = 4.0 moles H2SO4 / 1000 g solution

4.0 mol times 98.09 g/mol = 392.32 g

2) Determine equivalent weight for H2SO4:

98.09 g/mol / 2 dissociable hydrogen/mol = 49.05 g/equivalent

3) Determine # equivalents in 392.32 g:

392.32 g times (1 equivalent / 49.05 g) = 8.0 equivalents

4) Determine volume of solution:

392.32 g + 1000 g = 1392.32 g (total mass of solution)

1392.32 g / 1.2 g/mL = 1160.27 mL

5) Determine normality:

N = 8.0 equivalents / 1.16027 L = 6.9 N

Problem #13: An car antifreeze mixture is made by mixing equal volumes of ethylene glycol (d = 1.114 g/mL, molar mass 62.07 g/mol) and water (d = 1.000 g/mL) at 20.0 °C. The density of the solution is 1.070 g/mL.

Express the concentration of ethylene glycol as:

(a) volume percent
(b) mass percent
(c) molarity
(d) molality
(e) mole fraction

Solution to (a):

Since the volumes are equal, the volume percent of ethylene glycol is 50%

Solution to (b):

1) Determine the masses of equal volumes (we'll use 50.0 mL) of the two substances:

ethylene glycol: (50.0 mL) (1.114 g/mL) = 55.7 g
water: (50.0 mL) (1.000 g/mL) = 50.0 g

2) Determine percent due to ethylene glycol:

55.7 g / 105.7 g = 52.7%

Solution to (c):

1) Determine moles of ethylene glycol:

55.7 g / 62.07 g/mol = 0.89737 mol

2) Determine volume of solution:

105.7 g / 1.070 g/mL = 98.785 mL

3) Determine molarity:

0.89737 mol / 0.098785 L = 9.08 M

Solution to (d):

0.89737 mol / 0.050 kg = 17.9 m

Note the large difference between the molarity and the molality.

Solution to (e):

1) Determine moles of water:

50.0 g / 18.0 g/mol = 2.77778

2) Determine mole fraction for ethylene glycol:

0.89737 mol / 3.67515 mol = 0.244

Problem #14: What is the percent of CsCl by mass in a 0.0711 M CsCl solution that has a density of 1.09 g/mL?

Solution:

1) Determine mass of dissolved CsCl:

Let us assume 100.0 mL of solution.

MV = grams / molar mass

(0.0711 mol/L) (0.100 L) = x / 168.363 g/mol

x = 1.197 g

2) Determine mass of solution:

1.09 g/mL times 100.0 mL = 109 g

Determine mass percent of CsCl in solution:

1.197 g / 109 g = 1.098%

to three sig figs: 1.10%


Problem #15: A 8.77 M solution of an acid, HX, has a density of 0.853 g/mL.The acid, HX, has a molar mass of 31.00 g/mol. Determine the molal concentration of this solution, ΧHX (mole fraction of HX), and % w/w (percent by mass). The solvent in this solution is water, H2O.

Comment: Can an 8.77 M solution of an acid have a density of 0.853 g/mL? Who cares? We'll just solve the problem.

Solution:

There is a trick to solving this type of problem: let us assume 1.00 L (or 1000 mL) of the solution is present. (Another place where a similar trick is employed is in determining empirical formulas, where you assume 100 g of the substance is present.)

1) Some preliminary calculations:

moles acid: 1.00 L x (8.77 moles / L) = 8.77 moles HX (used in molality and mole fraction)

mass acid: 8.77 moles x 31.00 g / mole = 272 g HX (percent by mass)

mass of solution: 1000 mL x 0.853 g / mL = 853 g solution (percent by mass)

mass of solvent: 853 g minus 272 g = 581 g = 0.581 kg (molality)

moles solvent: 581 g divided by 18.015 g) = 32.25 (mole fraction)

2) Calculations to answer the questions:

molality = 8.77 moles / 0.581 kg = 15.1m

mole fraction = 8.77 / (8.77 + 32.25) = 0.214

% w/w = (272 g / 853 g) x 100 = 31.9%


Problem #16: A solution of hydrogen peroxide, H2O2, is 30.0% by mass and has a density of 1.11 g/cm3. Calculate the (a) molality, (b) molarity, and (c) mole fraction

Solution:

1) Mass of 1 liter of solution:

1.11 g/cm3 times (1000 cm3 / L) = 1110 g/L

2) Mass of the two components of the solution:

mass of H2O2 ---> 30.0% of 1110 g = 333 g
mass of H2O ---> 70.0% of 1110 g = 777 g

3) Moles of hydrogen peroxide:

333 g / 34.0138 g/mol = 9.79 mol

4) Molality:

9.79 mol / 0.777 kg = 12.6 m

5) Molarity

9.79 mol / 1.00 L = 9.79 M

6) Mole fraction

mole of H2O2 = 9.79
mole of H2O = 43.13

total moles ---> 9.79 + 43.13 = 52.92

mole fraction ---> 9.79 / 52.92 = 0.185


Problem #17: Household hydrogen peroxide is an aqueous solution containing 3.0% hydrogen peroxide by mass. What is the molarity of this solution? (Assume a density of 1.01 g/mL.)

Solution:

1) Let us have 1000 g of solution on hand. For the volume of solution:

1000 g divided by 1.01 g/mL = 990.1 mL

2) Since H2O2 is 3% by mass, we know that there are 30 grams of H2O2 present in the 1000 g of solution.

MV = mass / molar mass

(x) (0.9901 L) = 30 g / 34.0138 g/mol

x = 0.890814 M

Two sig figs seems reasonable, so 0.89 M.


Problem #18: A 6.90 M KOH solution in water has 30% by weight KOH. Calculate the density of the KOH solution.

Solution:

Let us assume we have 1.00 liter of solution present. This means we have 6.90 mol of KOH. Let us determine the mass of KOH:

6.90 mol times 56.1049 g/mol = 387.124 g

387.124 g represents 30% of the total weight of the solution. (The water makes up the other 70%.) To get the mass of the solution, do this:

387.124 g is to 0.3 as x is to 1

x = 1290 g

density of the solution ---> 1290 g / 1000 mL = 1.29 g/mL (to three sig figs)


Problem #19: An aqueous NaCl solution is made using 138 g of NaCl diluted to a total solution volume of 1.30 L.

(A) Calculate the molarity of the solution.
(B) Calculate the molality of the solution. (Assume a density of 1.08 g/mL for the solution.)
(C) Calculate the mass percent of the solution. (Assume a density of 1.08 g/mL for the solution.)

Solution:

Part A:

MV = mass / molar mass

(x) (1.30 L) = 138 g / 58.443 g/mol

x = 1.82 M

Part B:

molality is moles solute per kg of solvent. I will use 2.3613 mol (keeping a few guard digits).

Let us assume 1000 mL of the solution is present. This tells us that 2.3613 mol of the solute is present (that's the 138 g of NaCl).

1.08 g/mL times 1000 mL = 1080 g <--- this is the total mass of the 1000 mL solution

1080 g minus 138 g = 942 g <--- the mass of water in the 1000 mL of solution

942 g = 0.942 kg

2.3613 mol / 0.942 kg = 2.51 m

Part C:

138 g of solute was dissolved in 1080 total grams of solution

(138 / 1080) times 100 = 12.8% <--- NaCl

100% minus 12.8% = 87.2% <--- H2O

Another type of question is this area is to ask you to determine the mole fraction for each substance. For that you will need to know the moles of water:

942 g / 18.015 g/mol = 52.29 mol

The mole fraction of NaCl is this:

2.3613 mol / (2.3613 mol + 52.29 mol) = 0.0432

The mole fraction of the water is this:

1 - 0.0432 = 0.9568


Problem #20: A solution is prepared by dissolving 28.0 g of glucose (C6H12O6) in 350 g of water. The final volume of the solution is 384 mL . For this solution, calculate each of the following:

1) molarity; 2) molality; 3) percent by mass; 4) mole fraction; 5) mole percent

Solution:

1) Molarity is the number of moles divided by volume of solvent in liters.

28.0 g / 180 g/mol = 0.156 mol

0.156 mol / 0.384 L = 0.405 M

2) Molality is the number of moles divided by mass of solvent in kilograms.

0.156 mol / 0.350 kg = 0.444 m

3) Percent by mass, as the name implies, is the mass of solute divided by total mass times 100%.

28.0 g / (350 g + 28 g) times 100 = 7.41% C6H12O6 by mass

4) Mole fraction is the number of moles of solute divided by total moles.

350 g / 18.015 g/mol = 19.428 mol of water

0.156 mol / (19.428 + 0.156) = 0.0080 mol fraction C6H12O6

The mole fraction of water is:

1 - 0.0080 = 0.992

5) Mole percent is mole fraction times 100%.

0.0080 x 100 = 0.80 % C6H12O6 by moles

Problem #21: How many grams of glucose is necessary to dissolve in 3 litres of water to obtain 40% solution?

Solution:

Let x gms of glucose dissolve in 3 liters of water to form a 40% solution.

glucose weight = x
water weight = 3 liters = 3000 g (take water density as 1.00 g/mL)
so total solution weight = x + 3000 g

glucose percent ---> x / (x + 3000) = 0.40 <--- that's the 40%

x = 0.4 ( x + 3000)
x = 0.4x + 1200
x - 0.4x = 1200
0.6 x = 1200
x = 2000 g


Problem #22: The vinegar sold in the grocery stores is described as 5% (v/v) acetic acid. What is the molarity of this solution (density of 100% acetic acid is 1.05 g/mL)?

Solution:

1) 5% (v/v) means 5% of the volume is acetic acid. So 1.00 L of vinegar contains 50 mL acetic acid:

(0.05)(1000 mL) = 50 mL

2) The mass of this acetic acid is:

(1.05 g/mL)(50 mL) = 52.5 g

3) For the molarity, use MV = mass / molar mass

(x) (1.00 L) = 52.5 g / 60.0516 g/mol

x = 0.874 M (to three sig figs)


Problem #23: By titration, the molarity of acetic acid in vinegar was determined to be 0.870 M. Convert this to %(v/v). (The density of acetic acid is 1.05 g/mL)

Solution:

1) Assume 1.00 L of the solution to be present. Use MV = mass/molar mass to determine mass of acetic acid present:

(0.870 mol/L) (1.00 L) = x / 60.0516 g/mol

x = 52.245 g

2) Determine what volume of pure acetic acid this is:

52.245 g divided by 1.05 g/mL = 49.757 mL

3) Determine %(v/v):

(49.757 mL / 1000 mL) * 100 = 4.9757

Rounded off, this is 4.98%(v/v), usually given as 5%(v/v).


Problem #24: In an aqueous solution of sulfuric acid, the acid concentration is 2.40 mole percent and the density of the solution is 1.079 g/mL. Calculate (1) the molal concentration of the acid, (2) the weight percentage of the acid, and (3) the molarity of the solution

Solution:

2.40 mole percent of acid means 97.60 mole percent water.

Let's assume 100 moles of the solution is present. This means 2.40 mole of the solution is H2SO4 and 97.60 mole is water.

(1) For molality, we need to know kg of water ---> 97.60 mol times 18.015 g/mol = 1758.264 g = 1.758264 kg

molality ---> 2.40 mol / 1.758264 kg = 1.365 m (1.36 m to three sig figs)

(2) For the weight percent, we need the mass of H2SO4 ---> 2.40 mol times 98.0768 g/mol = 235.38432 g

weight percent ---> 235.38432 g / (235.38432 + 1758.264 g) = 0.118067 = 11.8% (three sig figs)

(3) For molarity, we need to know the volume of the solution ---> 1993.64832 g divided by 1.079 g/mL = 1847.68 mL = 1.84768 L

Note: 1993.64832 g is the total mass of the solution.

molarity ---> 2.40 mol / 1.84768 L = 1.2989 M (1.30 M to three sig figs)

By the way, mole percent is mole fraction written as a percent. The mole fractions in the above problem are 0.0240 and 0.9760.


Problem #25: Calculate the molality, molarity, and mole fraction of FeCl3 in a 26.3% (w/w) solution (density = 1.28 g/mL).

Solution:

Molarity:

Assume 100. g of solution present.

26.3 g of FeCl3 is present.

100. g divided by 1.28 g/mL = 78.125 mL

Use MV = mass / molar mass

(x) (0.078125 L) = 26.3 g / 162.204 g/mol

x = 2.08 M (to three sig figs)

Molality

100. g - 26.3 g = 73.7 g <--- the water in the 100. g of solution

molality ---> (26.3 g / 162.204 g/mol) / 0.0737 kg = 2.20 m (to three sig figs)

Mole fraction of FeCl3:

moles FeCl3 ---> 26.3 g / 162.204 g/mol = 0.1621415 mol

moles water ---> 73.7 g / 18.015 g/mol = 4.091035 mol

mole fraction ---> [0.1621415 mol / (0.1621415 mol + 4.091035 mol)] = 0.0381 (to three sig figs)


Problem #26: The mole fraction in a solution of Na2S is 0.125. Calculate the mass precent (w/w) of Na2S in this solution.

Solution:

The mole fraction of Na2S is 0.125. Therefore, the mole fraction of water in the solution is 0.875.

mass Na2S ---> 0.125 mol times 78.045 g/mol = 9.755625 g

mass water ---> 0.875 mole 18.015 g/mol = 15.763125 g % (w/w) Na2S ---> [9.755625 / (9.755625 + 15.763125) * 100] = 38.2% (to three sig figs)


Problem #27: In dilute nitric acid, the concentration of HNO3 is 6.00 M and the density of this solution is 1.19 g/mL. Use that information to calculate the mass percent and mole fraction of HNO3 in the solution.

Solution #1:

1) Let us assume 1000 mL of the solution is present. Some preliminary calculations:

1000 mL x 1.19 g/mL = 1190 g (this is the mass of our 1000 mL)

6.00 mol/L x 1.00 L = 6.00 mol (this is how many moles of HNO3 are in our 1000 mL)

6.00 mol x 63.012 g/mol = 378.072 g (the mass of HNO3 in our 1000 mL)

1190 g minus 378.072 g = 811.928 g (the mass of water in the 1000 mL of solution)

2) Calculate the mass percent of HNO3:

378.072 g / 1190 g = 31.8% (to three sig figs)

3) Calculate the mole fraction of HNO3:

811.928 g / 18.015 g/mol = 45.07 mol of H2O

6.00 mol / (6.00 mol + 45.07 mol) = 0.118 (to three sig figs)

Solution #2:

1) Let us assume 1000 g of the solution is present. Some preliminary calculations:

1000 g / 1.19 g/mL = 840.336 mL (the volume of our 1000 g of solution)

6.00 mol/L x 0.840336 L = 5.042 mol (this is how many moles of HNO3 are in our 840.336 mL)

5.042 mol x 63.012 g/mol = 317.7065 g (the mass of HNO3 in our 840.336 mL)

1000 g minus 317.7065 g = 682.2935 g (the mass of water in the 1000 g of solution)

2) Calculate the mass percent of HNO3:

317.7065 g / 1000 g = 31.8% (to three sig figs)

3) Calculate the mole fraction of HNO3:

682.2935 g / 18.015 g/mol = 37.874 mol of H2O

5.042 mol / (5.042 mol + 37.874 mol) = 0.117 (to three sig figs)

Comment: In solution #2, I obtained 0.117485 and that, technically, is 0.117 (not 0.118), when rounded to three sig figs.


Ten Examples

Problems 1 - 10

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