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The degree of dissociation of $Ca(NO_3)_2$ in a dilute aqueous solution containing 7g of salt per 100g of water at $100^{\large\circ}$C is 70%.Calculate the vapour pressure of solution.

$\begin{array}{1 1}746.24mm\\646.24mm\\546.24mm\\346.24mm\end{array} $

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$Ca(NO_3)_2\;\;\leftrightharpoons Ca^{2+}+2NO_3^-$
$\;\;1\qquad\qquad\qquad 0\qquad 0$
$1-\alpha\qquad\qquad\quad \alpha \qquad 2\alpha$
$\therefore$ Total moles at equilibrium =$1+2\alpha$
$\Rightarrow (1+2\times 0.7)=2.4$
$\alpha =0.7$
Also $m_N/m_{exp}=1+2\alpha$
$\therefore m_{exp}=\large\frac{m_N}{1+2\alpha}=\frac{164}{2.4}$$=68.33$
Further at $100^{\large\circ}C,P^0_{H_2O}$=760mm
$\large\frac{P^0-P_s}{P_s}=\frac{w\times M}{m\times W}$
$\therefore \large\frac{760-P_s}{P_s}=\frac{7\times 18}{68.33\times 100}$
answered Aug 7, 2014 by sreemathi.v

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