Q39 / 180 · KCET 2019 · Chemistry
A non-volatile solute, 'A' tetramerises in water to the extent of 80\%. \( 2.5 \mathrm{~g} \) of 'A' in \( 100 \mathrm{~g} \) of
water, lowers the freezing point by \( 0.3^{\circ} \mathrm{C} \). the molar mass of \( \mathrm{A} \) in \( \mathrm{g} \mathrm{mol}^{-1} \) is (Kf for water \( = \)
\( 1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \) )
water, lowers the freezing point by \( 0.3^{\circ} \mathrm{C} \). the molar mass of \( \mathrm{A} \) in \( \mathrm{g} \mathrm{mol}^{-1} \) is (Kf for water \( = \)
\( 1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \) )
- A\( 155 \)
- B\( 354 \)
- Correct answerC\( 62 \)
- D\( 221 \)
Solution
\( (C) \)
\[ \begin{array}{l} \alpha=\frac{80}{100}=0.8 \\ \alpha=\frac{1-i}{1-\frac{1}{n}}=\frac{1-i}{1-\frac{1}{4}} \\ 0.8=\frac{1-i}{\frac{3}{4}} \\ \Delta T_{f}=0.3 \\ \Delta T_{f}=\frac{i \times K_{f} \times W, \times 1000}{M_{2} \times 100} \\ M_{2}=\frac{0.4 \times 1.86 \times 2.5 \times 10}{0.3} \\ M_{2}=62 \mathrm{~g} m o l^{-1} \end{array} \]
\[ \begin{array}{l} \alpha=\frac{80}{100}=0.8 \\ \alpha=\frac{1-i}{1-\frac{1}{n}}=\frac{1-i}{1-\frac{1}{4}} \\ 0.8=\frac{1-i}{\frac{3}{4}} \\ \Delta T_{f}=0.3 \\ \Delta T_{f}=\frac{i \times K_{f} \times W, \times 1000}{M_{2} \times 100} \\ M_{2}=\frac{0.4 \times 1.86 \times 2.5 \times 10}{0.3} \\ M_{2}=62 \mathrm{~g} m o l^{-1} \end{array} \]
Hence, the correct answer is option: 'C'.