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Calculate the freezing point of an aqueous solution of a non-electrolyte having an osmotic pressure 2.0atm at 300K.$K_f=1.86Kmol^{-1}kg$ and $S=0.0821$litre atm $K^{-1}mol^{-1}$

$\begin{array}{1 1}-0.151^{\large\circ}C\\-0.251^{\large\circ}C\\-0.351^{\large\circ}C\\-0.101^{\large\circ}C\end{array} $

1 Answer

$\pi=2$ atm
$S=0.0821$ litre atm $k^{-1}mol^{-1}$
Also $\pi V=nST$
$\therefore$ Molarity =$\large\frac{n}{V}=\frac{\pi}{ST}$
$\Rightarrow \large\frac{2}{0.0821\times 300}$
$\Rightarrow$ 0.0812 mole litre$ ^{-1}$
For dilute solutions molarity =molality=0.0812
$\Delta T=K_f\times$ molality=$1.86\times 0.0812$
$\Rightarrow 0.151$
Freezing point =$T_0-\Delta T$
$\Rightarrow 0-0.151$
$\Rightarrow -0.151^{\large\circ}C$
answered Aug 7, 2014 by sreemathi.v

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