Q24 / 160 · TS EAMCET 2018 (05 May Shift 2) · Chemistry
$2.0 \mathrm{~g}$ of a non-electrolyte dissolved in $100 \mathrm{~g}$ of benzene lowers the freezing point of benzene by $1.2 \mathrm{~K}$. The freezing point depression constant of benzene is $5.12 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$. The molar mass of the solute is
- A$55 \mathrm{~g} \mathrm{~mol}^{-1}$
- Correct answerB$85 \mathrm{~g} \mathrm{~mol}^{-1}$
- C$120 \mathrm{~g} \mathrm{~mol}^{-1}$
- D$155 \mathrm{~g} \mathrm{~mol}^{-1}$
Solution
$$
\begin{aligned}
\Delta T_f & =K_f m=K_f \times \frac{W_2}{M_2} \times \frac{1000}{w_1} \\
1 \cdot 2 & =\frac{5 \cdot 12 \times 2 \times 1000}{M_2 \times 100} \\
M_2 & =85 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}
$$
Hence, the correct answer is option: 'B'.