A non-volatile non-electrolyte ( $5.0 g$ ) dissolved in $100 g$ of water (molar mass $18$ ) lowers the vapour pressure from $760 mmHg$ to $745 mmHg$ . Using Raoult's law exactly (no dilute-solution approximation), the molar mass of the solute in $g mo l^{- 1}$ is

Answer: B

p = x_{1} p^{0}

gives

(p^{0} - p) / p = n_{2} / n_{1}

exactly. So

(760 - 745) /745 = 0.02013 = \frac{5.0/ M}{100/18}

, giving

M = 5.0 \times 18/ (100 \times 0.02013) = 44.7

. The approximation

(p^{0} - p) / p^{0} \approx n_{2} / n_{1}

would give 45.6, which is not exact.