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JEE Main levelSingle correct optionDifficulty 2 of 3so060

When $3.0 g$ of a non-electrolyte is dissolved in $100 g$ of benzene, the freezing point falls by $1.28 K$ . Given $K_{f} (benzene) = 5.12 K kg mo l^{- 1}$ , the molar mass of the solute (in $g mo l^{- 1}$ ) is

A $12$
B $240$
C $120$
D $60$

Answer: C

M = \frac{1000 K _{f} w _{2}}{Δ T _{f} w _{1}} = \frac{1000 \times 5.12 \times 3.0}{1.28 \times 100} = 120 g mo l^{- 1}

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