Question

The degree of dissociation of $Ca(N{O}_3{)}_2$ in a dilute aqueous solution containing $7$ g of salt per $100$ g of water at ${100}^{o}C$ is $75\%$. The vapour pressure of solution in mm is $x$, then find the value of $x/150$ to the nearest integer.

Answer 1

$\underset{\underset{1-\alpha }{1}}{Ca(N{O}_3{)}_2}\rightleftharpoons \underset{\underset{\alpha }{0}}{C{a}^{2+}}+\underset{\underset{2\alpha }{0}}{2N{O}_3^{-}}$

$\therefore$ Total mole at equilibrium $=1+2\alpha$

$=(1+2\times 0.7)=2.4(\because \alpha =0.7)$

Also, $m_N/m_{exp}=1+2\alpha$

$\therefore m_{exp}=\frac{m_N}{1+2\alpha }=\frac{164}{2.4}=68.33$

Further at ${100}^{o}C,P_{H_{2}O}^0=760mm,w=7g,W=100g$

$\because \frac{P^0-P_S}{P_S}=\frac{w\times M}{m\times W}$

$\therefore \frac{760-P_S}{P_S}=\frac{7\times 18}{68.33\times 100}$

$\therefore P_S=746.26mm=x$

$x/150=4.97\approx 5$