Discussion
The freezing point of a mixture containing 1.60 g of naphthalene (molar mass = 128 g mol-1) and 20 g of benzene (molar mass = 78 g mol-1) is 2.8°C and that of pure benzene is 5.5°C. The value of the molal freezing point depression constant of benzene is :
Answer: A
Kf for benzene is 4.3°C kg mol-1.
Use ΔTf = Kf m, where m is the molality of naphthalene:
\[ \begin{aligned} \Delta T_{f} &= 5.5 - 2.8 = 2.7\,^{\circ}\text{C} \\ n &= 1.60/128 = 0.0125\ \text{mol} \\ m &= 0.0125/0.020 = 0.625\ \text{mol kg}^{-1} \end{aligned} \]
\[ \begin{aligned} K_{f} &= \frac{2.7}{0.625} \\ &= 4.32\,^{\circ}\text{C kg mol}^{-1} \end{aligned} \]
The unit follows from dividing kelvin (or °C) by mol kg-1, which gives °C kg mol-1. The options with g mol-1 or mol kg-1 have the right number but the wrong unit, and 5.1 would come from using benzene's molar mass where it does not belong.
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