The vapour pressure of water is $12.3 \, kPa$ at $300 \, K$. Calculate the vapour pressure of a $1 \, molal$ solution of a non-volatile solute in it.

(N/A) $1 \, molal$ solution means $1 \, mol$ of the solute is present in $1000 \, g$ of the solvent (water).
Molar mass of water $= 18 \, g \, mol^{-1}$.
Number of moles of water $= \frac{1000 \, g}{18 \, g \, mol^{-1}} = 55.56 \, mol$.
Mole fraction of the solute $(x_2)$ $= \frac{n_2}{n_1 + n_2} = \frac{1}{55.56 + 1} = \frac{1}{56.56} \approx 0.0177$.
Given,vapour pressure of pure water,$p_1^o = 12.3 \, kPa$.
Using Raoult's Law for non-volatile solute: $\frac{p_1^o - p_1}{p_1^o} = x_2$.
$p_1 = p_1^o (1 - x_2) = 12.3 \times (1 - 0.0177) = 12.3 \times 0.9823$.
$p_1 \approx 12.08 \, kPa$.