Question

$AB$ and $CD$ are two non-volatile solutes dissolved in same solvent, showing following behaviour
$AB\rightleftharpoons A^{+}+B^{-}$
$AB$ is ionized in solvent, $\alpha =$ degree of ionization
$CD\rightleftharpoons \frac{1}{2}(CD{)}_2$ $CD$ is dimerised in solvent, $\beta =$ degree of dimerisation
$\Delta T_b(AB)=2\Delta T_b(CD)$
If $\alpha =0.80$, then $\beta$ is

• A. 0.50
• B. 0.40
• C. 0.20
• D. 0.80

Answer 1

Answer: (C)

For $AB,i=1+(y-1)\alpha$

$y=$ number of particles from one mole of solute $=2$

$i=1+(2-1)\alpha =(1+\alpha )$

For CD, $i=\left(1+\left(\frac{1}{2}-1\right)\beta \right)=\left(1-\frac{\beta }{2}\right)=\left(\frac{2-\beta }{2}\right)$

$\therefore \Delta T_b(AB)=K_{b}m$ (molality) $i=K_{b}m(1+\alpha )$

$\Delta T_b(CD)=K_{b}mi=K_{b}m\left(\frac{2-\beta }{2}\right)$

Given $\frac{\Delta T_b(AB)}{\Delta T_b(CD)}=2$

$\therefore \frac{K_{b}m(1+\alpha )}{K_{b}m\left(\frac{2-\beta }{2}\right)}=2$

$\frac{1+\alpha }{(2-\beta )}=1$

$1+\alpha =2-\beta$

$\therefore \beta =1-\alpha =1-0.80=0.20$