Question

Calculate the percentage degree of dissociation of an electrolyte $X{Y}_2$ in water $\text{Normal molar mass}=164g/mol$, if the observed molar mass by measuring elevation in boiling point is $65.6g/mol$

• A. $75\%$
• B. $25\%$
• C. $60\%$
• D. $33\%$

Answer 1

The correct option is A $75\%$

$X{Y}_2\rightleftharpoons X^{2+}+2Y^{-}$
Initially $1$ $0$ $0$
At equilibrium $1-\alpha$ $\alpha$ $2\alpha$

$\text{Total no. of moles at equilibrum}=1-\alpha +\alpha +2\alpha =1+2\alpha$

$i=\frac{\text{Normal molar mass}}{\text{Observed molar mass}}=\frac{\text{Number of moles at equilibrium}}{\text{No. of moles initially}}\frac{1+2\alpha }{1}=\frac{164}{65.6}$
$\therefore \alpha =0.75$
% $\alpha =75$%