Question

$1.5\, g$ of a monobasic acid when dissolved in $150\, g$ of water lowers the freezing point by $0.165^{\circ} C .0 .5\, g$ of the same acid when titrated, after dissolution in water, requires $37.5 ml$ of $N / 10$ alkali. Calculate the degree of dissociation of the acid ($K_{f}$ for water $=1.86^{\circ} C\, mol ^{-1}$ ).

Answer 1

$\left(\frac{0.5}{ M }\right)=\left(\frac{37.5}{10}\right) \times 10^{-3}$
$M =\frac{5}{37.5} \times 1000$
$M =133.33\, gm$
$\Delta T _{ f }= iK_{f} \textrm {m }$
$0.165=\frac{ i \times 1.86 \times 1.5 / 133.3}{150 / 1000}$
$i =1.182$
$\alpha=\frac{ i -1}{ n -1} \{ n =2\}$
$\alpha=18.28 \%$