Q45 / 180 · KCET 2015 · Chemistry
After adding non-volatile solute freezing point of water decreases to \( -0.186^{\circ} \mathrm{C} . \) Calculate
\( \Delta T_{b} \) if \( K_{\mathrm{f}}=1.86 K \mathrm{~kg} \mathrm{~mol}^{-1} \) and \( K_{b}=0.521 K \mathrm{~kg} \mathrm{~mol}^{-1} \)
\( \Delta T_{b} \) if \( K_{\mathrm{f}}=1.86 K \mathrm{~kg} \mathrm{~mol}^{-1} \) and \( K_{b}=0.521 K \mathrm{~kg} \mathrm{~mol}^{-1} \)
- A\( 0.521 \mathrm{~K} \)
- Correct answerB\( 0.0521 \mathrm{~K} \)
- C\( 1.86 \mathrm{~K} \)
- D\( 0.0186 \mathrm{~K} \)
Solution
Depression of freezing point for the water is,
\[ \begin{array}{l} \Delta T_{f}=T_{f}^{\circ}-T_{f}=0-(-0.186)=0.186^{\circ} \mathrm{C} \\ \Delta T_{f}=K_{f} \times m \end{array} \]
Now, the molality of the solution,
\[ m=\frac{\Delta T_{f}}{K_{f}}=\frac{0.186}{1.86}=0.1 \mathrm{~m} \]
Elevation of boiling point for water is,
\[ \begin{array}{l} \Delta T_{b}=K_{b} \times m \\ =0.521 \times 0.1 \\ \Delta T_{b}=0.0521 K \end{array} \]
\[ \begin{array}{l} \Delta T_{f}=T_{f}^{\circ}-T_{f}=0-(-0.186)=0.186^{\circ} \mathrm{C} \\ \Delta T_{f}=K_{f} \times m \end{array} \]
Now, the molality of the solution,
\[ m=\frac{\Delta T_{f}}{K_{f}}=\frac{0.186}{1.86}=0.1 \mathrm{~m} \]
Elevation of boiling point for water is,
\[ \begin{array}{l} \Delta T_{b}=K_{b} \times m \\ =0.521 \times 0.1 \\ \Delta T_{b}=0.0521 K \end{array} \]
Hence, the correct answer is option: 'B'.