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Home  >>  CBSE XII  >>  Chemistry  >>  Solutions
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If boiling point of an aqueous solution is $100.1^{\large\circ}$C.What is its freezing point?Given,latent heat of fusion and vaporisation of water as 80 cal $g^{-1}$ and 540 cal $g^{-1}$ respectively.

$\begin{array}{1 1}-0.361^{\large\circ}C\\-0.461^{\large\circ}C\\-0.561^{\large\circ}C\\-0.161^{\large\circ}C\end{array} $

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Given : $L_f=80 cal g^{-1},L_{\nu}=540cal g^{-1}$
We know that,
$\Delta T_b=K_b\times $ molality
$\Delta T_f=K_f\times $ molality
Also,$K=\large\frac{RT^2}{1000\times L}$
$\therefore \large\frac{\Delta T_b}{\Delta T_f}=\frac{K_b}{K_f}$
$\Rightarrow \large\frac{RT_b^2}{1000\times L_{\nu}}\times \frac{1000\times L_f}{RT_f^2}$
$\large\frac{\Delta T_b}{\Delta T_f}=\frac{T_b^2\times L_f}{T_f^2\times L_{\nu}}$
$\large\frac{0.1}{\Delta T_f}=\frac{373\times 373\times 80}{273\times 273\times 540}$
$\therefore \Delta T_f=0.361$
Freezing point $(T_f)=0-\Delta T_f=0-0.361$
$\Rightarrow -0.361^{\large\circ}C$
answered Aug 7, 2014 by sreemathi.v
 

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