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When 5 g of non-volatile non-electrolyte solute is dissolved in 100 g of a certain solvent, the freezing point of the solvent decreases by 0.25 K. The molar mass of the solute is: [KfK_f of the given solvent = 1.2 K kg mol−11.2 \text{ K kg mol}^{-1}]

A

242.8 g mol−1242.8 \text{ g mol}^{-1}

B

238.2 g mol−1238.2 \text{ g mol}^{-1}

C

241.8 g mol−1241.8 \text{ g mol}^{-1}

D

240.0 g mol−1240.0 \text{ g mol}^{-1}

Step-by-Step Solution

The depression in freezing point (ΔTf\Delta T_f) is related to the molar mass of the solute (M2M_2) by the formula:

M2=Kf×w2×1000ΔTf×w1M_2 = \frac{K_f \times w_2 \times 1000}{\Delta T_f \times w_1}

Where: KfK_f = molal freezing point depression constant = 1.2 K kg mol−11.2 \text{ K kg mol}^{-1} w2w_2 = mass of solute = 5 g5 \text{ g} ΔTf\Delta T_f = depression in freezing point = 0.25 K0.25 \text{ K} w1w_1 = mass of solvent = 100 g100 \text{ g}

Substituting the values into the equation: M2=1.2×5×10000.25×100M_2 = \frac{1.2 \times 5 \times 1000}{0.25 \times 100} M2=600025=240.0 g mol−1M_2 = \frac{6000}{25} = 240.0 \text{ g mol}^{-1}

Therefore, the molar mass of the solute is 240.0 g mol−1240.0 \text{ g mol}^{-1}.

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