Question

The vapour pressure of a solution containing 2 g of an electrolyte BA in 100 g water which dissociates in one $B^{+}$ and one $A^{-}$ ion in water is 751 mm at ${100}^{o}C$. Percentage degree of ionisation of BA if its molar mass is 56 g $mo{l}^{-1}$ (write answer after dividing degree of dissociation by $80$ to the nearest integer) is_________.

Answer 1

$\frac{P^0-P_S}{P_S}=\frac{w\times M}{m\times W}$
$\therefore \frac{760-751}{751}=\frac{2\times 18}{m\times 100}$
$\therefore m_{exp}=30.04$ (for electrolyte BA)
Also, $\underset{\underset{1-\alpha }{1}}{BA}\rightleftharpoons \underset{\underset{\alpha }{0}}{B^{+}}+\underset{\underset{\alpha }{0}}{A^{-}}$
$\therefore m_N/m_{exp}=1+\alpha$
$\therefore 1+\alpha =56/30.04$ or $\alpha =0.8641$ or 86.41%

Answer is 1.