Question

0.150 mol of a solute is dissolved in 300. g of carbon tetrachloride. Calculate (a) the mole fraction and (b) the molality of solute.

Answer 1

a)

Molar mass of CCl4,

MM = 1*MM(C) + 4*MM(Cl)

= 1*12.01 + 4*35.45

= 153.81 g/mol

mass(CCl4)= 300 g

use:

number of mol of CCl4,

n = mass of CCl4/molar mass of CCl4

=(3*10^2 g)/(1.538*10^2 g/mol)

= 1.95 mol

Mole fraction of solute = mol of solute / total mol

= 0.150 / (0.150 + 1.95)

= 0.0714

Answer: 0.0714

b)

Mass of solvent = 300 g = 0.300 Kg

Use:

Molality = mol of solute / mass of solvent in Kg

= 0.150 mol / 0.300 Kg

= 0.500 m

Answer: 0.500 m