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Home  >>  CBSE XII  >>  Chemistry  >>  Solutions
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A 0.1539 molal aqueous solution of cane sugar (mol mass =$342gmol^{-1}$) has a freezing point of 271K while the freezing point of pure water is $273.15K$.What will be the freezing point of an aqueous solution containing 5g of glucose (mol mass =$180gmol^{-1}$) per 100g of solution.

$\begin{array}{1 1}269.27K\\367.29K\\169.27K\\469.27K\end{array} $

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1 Answer

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Molality =0.1539m
$\Delta T_f=273.15-271K$
$\qquad=2.15K$
$\Delta T_f=K_f.m$
$m=\large\frac{\Delta T_f}{K_f}$
$0.1539=\large\frac{2.15}{K_f}$
$K_f=\large\frac{2.15}{0.1539}$
$W_2=5g$
$M_2=180gmol^{-1}$
$\Delta T_f=K_f\times m$
$\Rightarrow \large\frac{2.15}{0.1539}\times \frac{5}{180}\times \frac{1000}{95}$
$\Rightarrow 4.08K$
Freezing point of solution =273.15-4.08K
$\Rightarrow 269.27K$
answered Aug 8, 2014 by sreemathi.v
 

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