The vapour pressure of pure benzene at a certain temperature is 0.85 bar. A non-volatile, non-electr — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

$P^\circ = 0.85 \text{ bar}, P = 0.845 \text{ bar}$. $\Delta P = 0.005 \text{ bar}$. Using $(P^\circ - P) / P = n / N$, $0.005 / 0.845 = (0.5 / M) / (39.0 / 78)$. Since $39/78 = 0.5$, we get $0.005 / 0.845 = (0.5 / M) / 0.5 = 1 / M$. Therefore, $M = 0.845 / 0.005 = 169 \text{ g/mol}$. Therefore, correct answer is A.