### MolalityProblems #1-15

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 solvent

(8.010 mol) (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 / 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 cm3

1.329 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) (1.000 L) = 15.00 mole of HCl

15.00 mol times 36.4609 g/mol = 546.9135 g of HCl

2) Use the density to get mass of solution

(1000. mL) (1.0745 g/cm3) = 1074.5 g of solution

1074.5 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: What is the mass of a sample of a 0.449 molal KBr that contains 2.92 kg of water?

Solution:

1) Molality = moles solute divided by kilograms solute:

0.449 mol/kg = x / 2.92 kg

x = 1.31108 mol of KBr

2) Moles times molar mass equals grams:

(1.31108 mol) (119.0023 g/mol) = 156 g KBr

156 g KBr + 2920 g water = 3076 g total

If you wanted to be real technical about it, then use three sig figs to obtain 3080 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) ( 86.1766 g/mol) = 33.6950 g

33.6950 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 tetrahydrofuran

(1.42 mol) ( 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) (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 kg

moles 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) (1.230 g/mL) = 1230 g

2) Determine mass of 3.75 mol of H2SO4:

(3.75 mol) (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) (1050 g/L) = 1050 g of solution.

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

(4.20 mol) (58.443 g/mol) = 245.4606 g <--- mass of NaCl in solution

1050 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) (120.921 g/mol) = 432.89718 g

1000 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

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) (1.59 g/mL) = 2.1306 g

2.1306 g / 153.823 g/mol = 0.013851 mol

2) Mass of the methylene chloride:

(65.0 mL) (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 mol

0.00581158 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 mol

mole 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 solution

0.825 is to 450.825 as x is to 1,000,000

x = 1830 ppm

Bonus Problem: 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 #1:

1) Molality is moles solute dissolved per kilogram of solvent.

 mol solute m = ––––––––– kg solvent

2) Let moles of solute be represented by 'n.'

 mass solute n = ––––––––––––––––– molar mass of solute

3) The formula for acetone is C3H6O and its molar mass is 58.0794 g/mol, which equals 0.0580794 kg/mol.

I'm going to use the kg/mol amount and the reason will show up in a moment.

 mass solute n = –––––––––––– 0.0580794 kg/mol

4) Given the following:

mass of solution = mass of solvent + mass of solute = 450.0 g = 0.4500 kg

I'll ignore those two trailing zeros for the moment.

We write this:

mass of solvent = 0.45 − mass of solute

Here's the reason why I must use the kg/mol unit: In the subtraction just above, the 0.45 is in kilograms. That means the mass of solute must also be in kilograms. You can't subtract two numbers using different units.

Also, the bottom unit must be in kilograms because the 0.75 molal value is determined with kg in the denominator. Using grams in the denominator is not done with molality.

5) We are now ready to solve the problem. Let the mass of solute (in kilograms) be represented by 'x.'

 x –––––– 0.0580794
0.75 = ––––––––––––
0.45 − x

 x 0.3375 − 0.75x = ––––––––– 0.0580794

x = 0.019602 − 0.04356x

1.04356x = 0.019602

x = 0.01878 kg = 18.78 g (to 4 sig figs)

Solution #2:

0.7500 molal means 0.7500 mole of solute (the acetone) per 1000 g of water

mass 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