Question

When 11.6 g of an unknown, non-volatile, non-electrolyte, X was dissolved in 100. g of benzene, the vapor pressure of the solvent decreased from 100 torr to 90.3 torr at 299 K. Calculate the molar mass of the solute, X.

Answer 1

According to Raoult’s law:

P = Po*X(solvent)

90.3 = 100 * X(solvent)

X(solvent) = 0.903

this is mole fraction of benzene

Molar mass of C6H6,

MM = 6*MM(C) + 6*MM(H)

= 6*12.01 + 6*1.008

= 78.11 g/mol

mass(C6H6)= 100 g

number of mol of C6H6,

n = mass/molar mass

=(100.0 g)/(78.11 g/mol)

= 1.28 mol

X(benzene) = n(benzene) / (n(benzene) + n(unknown))

0.903 = 1.28 / (1.28 + n(unknown))

1.156 + 0.903*n(unknown) = 1.28

0.903*n(unknown) = 0.124

n(unknown) = 0.137

This is number of mol of non volatile

use:

n(unknown) = mass /molar mass

0.137 = 11.6 / MM

MM = 84.5 g/mol

Answer: 84.5 g/mol