Question

A solution was prepared by dissolving 39.0 g of KCl in 225 g of water.

Part A:

Calculate the mole fraction of the ionic species KCl in the solution.

Express the concentration numerically as a mole fraction in decimal form.

Note: The answer is not 0.0419..

Part B:

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

Express your answer with the appropriate units.

Part C:

Calculate the molality of KCl in the solution.

Express your answer with the appropriate units.

Answer 1

PART-A

Mole fraction of the ionic species KCl in the solution:

Total mass of solution = Mass of Solute + Mass of Solvent.

Total mass of solution = 39 g + 225 g = 264 g

Moles of KCl = wt/ mol wt = 39 g / 74.55 g/mol = 0.523 moles

Moles of water = wt/ mol wt = 225 g / 18 g/mol = 12.500 moles

Total moles of solution = 0.523 moles +  12.50 moles = 13.023 moles

Mole fraction Cl⁻ = Moles Cl⁻ / Total moles of solution

Mole fraction K⁺ = Moles K⁺ / Total moles of solution

In both cases (Mole fraction Cl⁻ and Mole fraction K⁺) = 0.523 / 13.023 = 0.040

Part B:Calculate the molarity of KCl

Molarity of KCl is the moles of solute (KCl) per litre of solution

volume of solution = 239 mL = 0.239 L

Molarity of KCl = moles of KCl / vol of sol in L = 0.523 moles / 0.239 L = 2.19 mol/L = 2.19M

Molarity of KCl = 2.19M

Part C:Calculate the molality of KCl

Molality of KCl is the moles of solute per kg of solvent.

Mass of solvent = 225 g = 0.225 kg

molality of KCl  = moles of KCl / mass of solvent in kg

molality of KCl  = 0.523 moles / 0.225 kg = 2.32 mol/kg = 2.32 molal

molality of KCl = 2.32 molal