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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 vaporization of water as $80cal g^{-1}$ and $540cal g^{-1}$ respectively.

$\begin{array}{1 1}-0.361^{\large\circ}C\\-0.461^{\large\circ}C\\-0.561^{\large\circ}C\\-0.661^{\large\circ}C\end{array} $

1 Answer

Given :
$L_f=80cal g^{-1}$
$L_{\nu}=540cal g^{-1}$
We know that
$\Delta T_b=K_b\times$ molality
$\Delta T_f=K_f\times$ molality
Also,$K=\large\frac{RT^2}{1000\times L}$
$\therefore \large\frac{\Delta T_b}{\Delta T_f}=\frac{K_b}{K_f}$
$\Rightarrow\large\frac{RT_b^2}{1000\times L_{\nu}}\times \frac{1000\times L_f}{RT_f^2}$
(Or) $\large\frac{\Delta T_b}{\Delta T_f}=\frac{T_b^2\times L_f}{T_f^2\times L_{\nu}}$
(Or) $\large\frac{0.1}{\Delta T_f}=\frac{373\times 373\times 80}{273\times 273\times 540}$
$\therefore \Delta T_f=0.361$
Freezing point $(T_f)=0-\Delta T_f$
answered Aug 6, 2014 by sreemathi.v

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