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Exactly of an unknown, non-electrolyte solute dissolved in of benzene lowered the freezing point of Solutions and Colligative Properties Chemistry Question

Question

Exactly $1.00\text{ g}$ of an unknown, non-electrolyte solute dissolved in $50\text{ g}$ of benzene lowered the freezing point of the benzene by $0.40\text{ K}$. If the $K_f$ of benzene is $5.12\text{ K kg mol}^{-1}$, calculate the experimental molar mass of the solute.

Answer: B

💡 Solution & Explanation

Using the depression in freezing point formula: $\Delta T_f = K_f \times \frac{w_{solute} \times 1000}{M_{solute} \times W_{solvent}}$. Rearranging algebraically to solve for $M_{solute}$ gives $M_{solute} = \frac{5.12 \times 1.00 \times 1000}{0.40 \times 50} = \frac{5120}{20} = 256\text{ g mol}^{-1}$.

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