The boiling point of aqueous glucose solution is $101.3 {}^{o}C$ . Its freezing point is (Given that $K_{f}$ of water $= 1.86 {}^{o}C / m$ and $K_{b}$ of water $= 0.52 {}^{o}C / m$ )

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$- {2.65}^{\circ} C$

$- {4.65}^{\circ} C$

${2.65}^{\circ} C$

${4.65}^{\circ} C$

answer is B.

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Detailed Solution

$\frac{∆ T_{b}}{∆ T_{f}} = \frac{K_{b}}{K_{f}} \frac{T_{s} - T_{b}}{T_{0} - T_{f}} = \frac{K_{b}}{K_{f}} \frac{101.3 - 100}{0 - T_{f}} = \frac{0.52}{1.86} T_{f} = - \frac{1.3}{0.279} = - {4.65}^{\circ} C$

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