The latent heat of vaporization of a liquid of molar mass, 80 g/mol and boiling point, is 8 kcal/mol — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

$K_b = (R \cdot T_b^2 \cdot M) / (1000 \cdot \Delta H_{\text{vap}})$. $T_b = 127 + 273 = 400 \text{ K}$. $M = 80 \text{ g/mol} = 0.08 \text{ kg/mol}$. $R \approx 2 \text{ cal mol}^{-1}\text{K}^{-1}$. $\Delta H_{\text{vap}} = 8000 \text{ cal/mol}$. $K_b = (2 \times (400)^2 \times 80) / (1000 \times 8000) = (2 \times 160000 \times 80) / 8000000 = 25600000 / 8000000 = 3.2 \text{ K-kg/mol}$. Therefore, correct answer is A.