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) $i=K_{b} m(1+\alpha)$

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

Given $\frac{\Delta T_{b}(A B)}{\Delta T_{b}(C D)}=2$

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

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

$\quad 1+\alpha=2-\beta$

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