Chemistry - Liquid Solution Question with Solution | TestHub

Given:
Mass of solute, $w_2=13$ g
Mass of solvent (water), $w_1=100$ g
Vapour pressure of pure water, $P^0=28.065$ mm Hg
Vapour pressure of solution, $P_s=27.371$ mm Hg
Molar mass of water, $M_1=18$ g/mol

Using the formula for relative lowering of vapour pressure (assuming dilute solution):
$\frac{P^0-P_s}{P^0}=\frac{n_2}{n_1}$

$\frac{P^0-P_s}{P^0}=\frac{w_2\times M_1}{M_2\times w_1}$

$M_2=\frac{w_2\times M_1\times P^0}{w_1\times (P^0-P_s)}$

Substituting the given values:
$M_2=\frac{13\times 18\times 28.065}{100\times (28.065-27.371)}$

$M_2=\frac{13\times 18\times 28.065}{100\times 0.694}$

$M_2\approx 94.74$ g/mol

Answer: 94.74 g/mol