Problems #11 - 25

**Problem #11:** Calculate the molarity and mole fraction of acetone in a 2.28-molal solution of acetone (CH_{3}COCH_{3}) in ethanol (C_{2}H_{5}OH). (Density of acetone = 0.788 g/cm^{3}; density of ethanol = 0.789 g/cm^{3}.) 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 g132.421 g divided by 0.788 g/cm

^{3}= 168.047 cm^{3}

ethanol

1000 g divided by 0.789 g/cm^{3}= 1267.427 cm^{3}

total volume

168.047 + 1267.427 = 1435.474 cm^{3}

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 H_{2}SO_{4} present:

4.0 molal = 4.0 moles H_{2}SO_{4}/ 1000 g solution4.0 mol times 98.09 g/mol = 392.32 g

2) Determine equivalent weight for H_{2}SO_{4}:

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 mNote 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, Χ_{HX} (mole fraction of HX), and % w/w (percent by mass). The solvent in this solution is water, H_{2}O.

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 = 15.1mmole 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, H_{2}O_{2}, is 30.0% by mass and has a density of 1.11 g/cm^{3}. Calculate the (a) molality, (b) molarity, and (c) mole fraction

**Solution:**

1) Mass of 1 liter of solution:

1.11 g/cm^{3}times (1000 cm^{3}/ L) = 1110 g/L

2) Mass of the two components of the solution:

mass of H_{2}O_{2}---> 30.0% of 1110 g = 333 g

mass of H_{2}O ---> 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 H_{2}O_{2}= 9.79

mole of H_{2}O = 43.13total 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 H_{2}O_{2} is 3% by mass, we know that there are 30 grams of H_{2}O_{2} 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 1x = 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% <--- H

_{2}O

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 molThe 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 (C_{6}H_{12}O_{6}) 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 mol0.156 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% C_{6}H_{12}O_{6}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 water0.156 mol / (19.428 + 0.156) = 0.0080 mol fraction C

_{6}H_{12}O_{6}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 % C_{6}H_{12}O_{6}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 gglucose percent ---> x / (x + 3000) = 0.40 <--- that's the 40%

x = 0.4 (x + 3000)

x = 0.4x + 1200

x − 0.4x = 1200

0.6x = 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/molx = 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/molx = 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 = 4.9757Rounded 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 H

_{2}SO_{4}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 H

_{2}SO_{4}---> 2.40 mol times 98.0768 g/mol = 235.38432 gweight 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 FeCl_{3} in a 26.3% (w/w) solution (density = 1.28 g/mL).

**Solution:**

Molarity:

Assume 100. g of solution present.26.3 g of FeCl

_{3}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 solutionmolality ---> (26.3 g / 162.204 g/mol) / 0.0737 kg = 2.20 m (to three sig figs)

Mole fraction of FeCl_{3}:

moles FeCl_{3}---> 26.3 g / 162.204 g/mol = 0.1621415 molmoles 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 Na_{2}S is 0.125. Calculate the mass precent (w/w) of Na_{2}S in this solution.

**Solution:**

The mole fraction of Na_{2}S is 0.125. Therefore, the mole fraction of water in the solution is 0.875.mass Na

_{2}S ---> 0.125 mol times 78.045 g/mol = 9.755625 gmass water ---> 0.875 mole 18.015 g/mol = 15.763125 g % (w/w) Na

_{2}S ---> [9.755625 / (9.755625 + 15.763125) * 100] = 38.2% (to three sig figs)

**Problem #27:** In dilute nitric acid, the concentration of HNO_{3} 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 HNO_{3} 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 HNO

_{3}are in our 1000 mL)6.00 mol x 63.012 g/mol = 378.072 g (the mass of HNO

_{3}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 HNO_{3}:

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

3) Calculate the mole fraction of HNO_{3}:

811.928 g / 18.015 g/mol = 45.07 mol of H_{2}O6.00 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 HNO

_{3}are in our 840.336 mL)5.042 mol x 63.012 g/mol = 317.7065 g (the mass of HNO

_{3}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 HNO_{3}:

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

3) Calculate the mole fraction of HNO_{3}:

682.2935 g / 18.015 g/mol = 37.874 mol of H_{2}O5.042 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.

**Problem #28:** A 0.100 M NaOH solution will be prepared by dilution of a 50.0% (w/w) NaOH solution. This solution has a density of 1.53 g/mL. Compute the volume of this solution that is required to prepare 1.00 x 10^{3} mL of 0.100 M NaOH.

**Solution:**

1) Determine moles of NaOH in 1.0 x 10^{3} mL of 0.10 M solution:

MV = moles(0.100 mol/L) (1.00 L) = 0.100 mol

2) Determine mass of 0.100 mol of NaOH

moles x molar mass = grams(0.100 mol) (40.00 g/mol) = 4.00 g

3) Mass of 50.0% (w/w) solution that contains 4.00 g of NaOH:

use a ratio and proportion50 g is to 100 g as 4 g is to x

x = 8.00 mL

4) Determine volume of solution that contains 8.00 g of NaOH:

Luke, uuuuuuse the density!8.00 g / 1.53 g/mL = 5.23 mL

**Problem #29:** What is the molarity of a NaOH solution with a density of 1.33 g/mL that was made with 70.0 ml of water if the molality is 10.7 molal?

**Solution:**

1) Use the molality to determine how much NaOH was used with the 70.0 g of water:

10.7 molal = 10.7 mol solute per kg solvent10.7 mol/kg = x / 0.0700 kg

x = 0.749 mol of NaOH

2) Determine the grams of NaOH:

(0.749 mol) (40.0 g/mol) = 29.96 g <--- gonna use 30.0

3) Total mass of the solution:

70.0 g + 30.0 g = 100.0 g

4) Volume of the solution:

100.0 g / 1.33 g/mL = 75.2 mL

5) Molarity:

0.749 mol / 0.0752 L = 9.96 M

**Problem #30a:** How many mL of an 3.78% (w/w) solution can be prepared from 18.00 g of sucrose?

**Solution:**

3.78 is to 18 as 100 is to xx = 476.19 g <--- total mass of solution

476.19 − 18.00 = 458.19 g <--- mass of water

Assume 1 g/mL for the density and no volume change when combining 458.19 g of water and 18.00 g of sucrose.

Answer = 458 mL (to three sig figs)

**Problem #30b:** How many mL of an 3.78% (w/v) solution can be prepared from 18.00 g of sucrose?

**Solution:**

3.78%(w/v) means 3.78 g of sucrose per 100 mL of solution3.78 is to 100 as 18 is to x

x = 476 mL (to three sig figs)