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Standard XII

Chemistry

Depression in Freezing Point

A solution of...

Question

A solution of urea in water has a boiling point of 100.18 $^{o}$ C. Calculate the freezing point of the solution. (K $_{f}$ for water is 1.86 K kg mol $^{- 1}$ and K $_{b}$ for water is 0.512 K kg mol $^{- 1}$ ).

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Solution

Given : Boiling point of solution $= {100.18}^{o} C$

$K_{b} = 0.512 K k g m o l^{- 1}$

$K_{f} = 1.86 K k g m o l^{- 1}$

$Δ T$ Boiling $= {100.18}^{o} - 100^{o} = 0.18$

$Δ T = K_{b} \times m o l a l i t y$

$∴$ molality $= \frac{0.18}{0.512}$

Now, $Δ T_{f r e e z i n g} = K_{f} \times m o l a l i t y$
$= 1.86 \times \frac{0.18}{0.512} = 0.6539$

$∴$ Freezing point of solution $= 0 - {0.6539}^{o} C$
$= - {0.6539}^{o} C$

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A solution of urea in water has a boiling point of 100.18oC. Calculate the freezing point of the solution. (Kf for water is 1.86 K kg mol−1 and Kb for water is 0.512 K kg mol−1).

Given : Boiling point of solution =100.18oCKb=0.512Kkgmol−1Kf=1.86Kkgmol−1ΔT Boiling =100.18o−100o=0.18ΔT=Kb×molality∴ molality =0.180.512Now, ΔTfreezing=Kf×molality =1.86×0.180.512=0.6539∴ Freezing point of solution =0−0.6539oC =−0.6539oC

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