The molar latent heat of vaporization of a certain liquid is 4.9 kcal/mol. The normal boiling point — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

According to the Clausius-Clapeyron equation, $dP/dT = (P \Delta H_{vap}) / (R T^2)$. We need $dT/dP = (R T^2) / (P \Delta H_{vap})$. Boiling point $T = 77^\circ C = 350$ K. $\Delta H_{vap} = 4900$ cal/mol. $R \approx 2$ cal/(mol K). $P = 1$ atm. $dT/dP = (2 \times (350)^2) / (1 \times 4900) = (2 \times 122500) / 4900 = 245000 / 4900 = 50$ deg/atm. Therefore, correct answer is A.