13 g nonvolatile solute dissolved in 100 gm of water gives vapour...

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):
$\dfrac{P^0 - P_s}{P^0} = \dfrac{n_2}{n_1}$

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

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

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

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

$M_2 \approx 94.74$ g/mol

Answer: 94.74 g/mol