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Standard XII

Chemistry

Depression in Freezing Point

A solution of...

Question

A solution of urea in water has boiling point of ${100.15}^{o} C .$ Calculate the freezing point of the same solution if $K_{f}$ and $K_{b}$ for water are 1.87 K kg $m o l^{- 1}$ and 0.52 K kg $m o l^{- 1}$ respectively.

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Solution

$△ T_{b} = (100.15 - 100) = {0.15}^{o} C$

We know that, $△ T_{b} = m o l a l i t y \times K_{b}$

Molality $= \frac{△ T_{b}}{K_{b}} = \frac{0.15}{0.52} = 0.2884$

$△ T_{f} = m o l a l i t y \times K_{f}$

$= 0.2884 \times 1.87 = {0.54}^{o} C$

Thus, the freezing point of the solution $= 0 - {0.54}^{o} C$ $= - {0.54}^{o} C$

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A solution of urea in water has boiling point of 100.15oC. Calculate the freezing point of the same solution if Kf and Kb for water are 1.87 K kg mol−1 and 0.52 K kg mol−1 respectively.

△Tb=(100.15−100)=0.15oCWe know that, △Tb=molality×KbMolality =△TbKb=0.150.52=0.2884△Tf=molality×Kf =0.2884×1.87=0.54oCThus, the freezing point of the solution =0−0.54oC=−0.54oC

A solution of urea in water has boiling point of ${100.15}^{o} C .$ Calculate the freezing point of the same solution if $K_{f}$ and $K_{b}$ for water are 1.87 K kg $m o l^{- 1}$ and 0.52 K kg $m o l^{- 1}$ respectively.