Discussion
When 1 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.20 K. The freezing point depression constant of benzene is 5 K kg mol-1 . The molar mass (g /mol) of the solute is
Answer: A
The molar mass is 500 g/mol.
Freezing point depression: \(\Delta T_{f} = K_{f}m\), where the molality \(m = \frac{w_{2}/M_{2}}{w_{1}/1000}\):
\[ \begin{aligned} M_{2} &= \frac{K_{f} \times w_{2} \times 1000}{\Delta T_{f} \times w_{1}} \\ &= \frac{5 \times 1 \times 1000}{0.20 \times 50} = 500 \end{aligned} \]
The common slips are forgetting the 1000 that converts grams of benzene to kilograms, which gives 0.5, or using the masses the wrong way round. The other options do not satisfy the equation with these data.
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