Explanation

(c)
Given,
Weight of non-volatile solute,
w= 25 g
Weight of solvent, W=100 g
Lowering of vapour pressure,
$p^{{^\circ}} - p_{s} = 0.225\ mm\$
Vapour pressure of pure solvent,
$p^{{^\circ}} = 17.5\ mm\$
Molecular weight of solvent ($H_{2}O$), M =18 g
Molecular weight of solute, m=?
According to Raoult’s law
$\frac{p^{{^\circ}} - p_{s}}{p^{{^\circ}}} = \frac{w \times M}{m \times W}\ \ \ \$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{0.225}{17.5\ } = \frac{25 \times 18}{m \times 100}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ m = \frac{25 \times 18 \times 17.5}{22.5\ }\$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 350\ g$