A very small amount of non-volatile solute (that does not dissociate) is dissolved in of benzene (de — Solutions and Colligative Properties Chemistry Question
Question
A very small amount of non-volatile solute (that does not dissociate) is dissolved in $56.8 \text{ cm}^3$ of benzene (density $0.889 \text{ g cm}^{-3}$). At room temperature, vapour pressure of this solution is 100 mm Hg while that of benzene is 102 mm Hg. If the freezing temperature of this solution is 1.3 degree lower than that of benzene, what is the value of molal freezing point depression constant of benzene?
💡 Solution & Explanation
Mass of benzene $= 56.8 \times 0.889 = 50.495 \text{ g}$. Moles of benzene $N = 50.495 / 78 \approx 0.6474 \text{ mol}$. From Raoult's law, $(P^\circ - P) / P = n / N \Rightarrow 2 / 100 = n / 0.6474 \Rightarrow n = 0.01295 \text{ mol}$. Molality $m = n / W_{\text{kg}} = 0.01295 / 0.050495 = 0.256 \text{ m}$. Using $\Delta T_f = K_f \times m$, we have $1.3 = K_f \times 0.256 \Rightarrow K_f = 1.3 / 0.256 \approx 5.07 \text{ deg/molal}$. Therefore, correct answer is A.