Question

A solution was prepared by dissolving 31.0 g of KCl in 225 g of water.

Part A: Calculate the mass percent of KCl in the solution.

Part B: Calculate the mole fraction of the ionic species KCl in the solution.

Part C: Calculate the molarity of KCl in the solution if the total volume of the solution is 239 mL.

Part D: Calculate the molality of KCl in the solution.

Answer 1

A solution was prepared by dissolving 31.0 g of KCl in 225 g of water.

Part A: Calculate the mass percent of KCl in the solution.

% m/m = mass of KCl / mass of solution * 100%

mass of solution = mass of KCl + mass of water = 31+225 = 256

% m/m = 31 / 256 * 100 = 12.10 %

Part B: Calculate the mole fraction of the ionic species KCl in the solution.

mol frac. of K+ = mol of K+ / (Total mol)

mol frac. of Cl- = mol of Cl- / (Total mol)

mol frac. of H2O = mol of H2O / (Total mol)

mol of KCl = mass/MW = 31/74.5513 = 0.41582 mol of KCl

mol ofK+ = mol of Cl- = mol of KCl = 0.41582

mol of water = mass/MW = 225/18 = 12.5

now...

total mol = 0.41582 + 0.41582 + 12.5 = 13.33164

mol frac K+ = (0.41582 )/13.33164 = 0.03119 = 0.032

mol frac. of Cl- =0.41582 / (13.33164 ) = 0.03119 = 0.032

mol frac. of H2O =12.5  / (13.33164 ) = 0.9376

Part C: Calculate the molarity of KCl in the solution if the total volume of the solution is 239 mL.

M = mol of KCl / Vtotal

V total = 239 mL = 0.239 L

mol of KCl = mass/MW = 31/74.5513 = 0.41582 mol of KCl

M = 0.41582 /0.239 = 1.73983 M

Part D: Calculate the molality of KCl in the solution.

Molality = mol of solute / kg of solvent

kg solvent = 225 g = 0.225 kg

mol = 0.41582

molal = 0.41582 /0.225 = 1.8480