Question

A solution was prepared by dissolving 26.0 g of KCl in 225 g of water.

Part A:

Calculate the mole fraction of KCl in the solution.

Part B:

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

Part C:

Calculate the molality of KCl in the solution.

Answer 1

A)

Molar mass of KCl,

MM = 1*MM(K) + 1*MM(Cl)

= 1*39.1 + 1*35.45

= 74.55 g/mol

mass(KCl)= 26 g

use:

number of mol of KCl,

n = mass of KCl/molar mass of KCl

=(26 g)/(74.55 g/mol)

= 0.3488 mol

Molar mass of H2O,

MM = 2*MM(H) + 1*MM(O)

= 2*1.008 + 1*16.0

= 18.016 g/mol

mass(H2O)= 225 g

use:

number of mol of H2O,

n = mass of H2O/molar mass of H2O

=(2.25*10^2 g)/(18.02 g/mol)

= 12.49 mol

Mole fraction of KCl = mole of KCl / total number of mol

= 0.3488 / (0.3488 + 12.49)

= 0.0272

Answer: 0.0272

B)

volume , V = 2.39*10^2 mL

= 0.239 L

use:

Molarity,

M = number of mol of KCl / volume in L

= 0.3488/0.239

= 1.459 M

Answer: 1.46 M

C)

m(solvent)= 225 g

= 0.225 kg

use:

Molality,

m = number of mol of KCl / mass of solvent in Kg

=(0.3488 mol)/(0.225 Kg)

= 1.55 molal

Answer: 1.55 molal