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1.00 g of a non-electrolyte dissolved in 50 g of benzene lowers its freezing point by 0.40 K. Given Kf = 5.12 K kg/mol for benzene, the molar mass of the solute is:
A128 g/mol
B64 g/mol
C256 g/mol
D512 g/mol
Answer & Solution
Correct answer: C. 256 g/mol
1. $M_2 = (K_f \times w_2 \times 1000)/(\Delta T_f \times w_1)$.
2. = $(5.12 \times 1.00 \times 1000)/(0.40 \times 50)$.
3. = $5120 / 20$.
4. = $256$ g/mol. (Option A would come from doubling $\Delta T_f$.)
_Source: NCERT Class 12 Chemistry Ch 1 "Solutions", p.19_
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