KCET · Chemistry · Solutions
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 \( = \)
\( 1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \) )
- A \( 155 \)
- B \( 354 \)
- C \( 62 \)
- D \( 221 \)
Answer & Solution
Correct Answer
(C) \( 62 \)
Step-by-step Solution
Detailed explanation
\( (C) \)
\[
\begin{array}{l}
\alpha=\frac{80}{100}=0.8 \\
\alpha=\frac{1-i}{1-\frac{1}{n}}=\frac{1-i}{1-\frac{1}{4}} \\
0.8=\frac{1-i}{\frac{3}{4}} \\
\Delta T_{f}=0.3 \\
\Delta T_{f}=\frac{i \times K_{f} \times W, \times 1000}{M_{2} \times 100} \\
M_{2}=\frac{0.4 \times 1.86 \times 2.5 \times 10}{0.3} \\
M_{2}=62 \mathrm{~g} m o l^{-1}
\end{array}
\]
\[
\begin{array}{l}
\alpha=\frac{80}{100}=0.8 \\
\alpha=\frac{1-i}{1-\frac{1}{n}}=\frac{1-i}{1-\frac{1}{4}} \\
0.8=\frac{1-i}{\frac{3}{4}} \\
\Delta T_{f}=0.3 \\
\Delta T_{f}=\frac{i \times K_{f} \times W, \times 1000}{M_{2} \times 100} \\
M_{2}=\frac{0.4 \times 1.86 \times 2.5 \times 10}{0.3} \\
M_{2}=62 \mathrm{~g} m o l^{-1}
\end{array}
\]
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