Question

Calculate the percentage degree of dissociation of an electrolyte $X{Y}_2$ (Normal molar mass = $164$) in water if the observed molar mass by measuring elevation in boiling point is $65.6$

Answer 1

$\begin{array}{cccccc}& X{Y}_2& \rightleftharpoons & X^{2+}& +& 2Y^{-}\\ \text{Initially}& 1& & 0& & 0\\ \text{at equilibrium}& 1-\alpha & & \alpha & & 2\alpha \end{array}$
$\text{Total no. of moles}=1-\alpha +\alpha +2\alpha =1+2\alpha$
$i=\frac{\text{Normal molar mass}}{\text{Observed molar mass}}=\frac{1+2\alpha }{1}=\frac{164}{65.6}$
$\therefore \alpha =0.75;$ %$\alpha =0.75\times 100=75$%