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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 \( K_{b}=0.521 K \mathrm{~kg} \mathrm{~mol}^{-1} \)
Options:
\( \Delta T_{b} \) if \( K_{\mathrm{f}}=1.86 K \mathrm{~kg} \mathrm{~mol}^{-1} \) and \( K_{b}=0.521 K \mathrm{~kg} \mathrm{~mol}^{-1} \)
Solution:
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Verified Answer
The correct answer is:
\( 0.0521 \mathrm{~K} \)
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,
\[
\begin{array}{l}
\Delta T_{b}=K_{b} \times m \\
=0.521 \times 0.1 \\
\Delta T_{b}=0.0521 K
\end{array}
\]
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