### MolalityProblems #11-25 (incomplete)

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 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 times 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 times 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 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 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

Problem #16: 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

3) Add 'em up:

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.