Question

$10g$ of ${NH}_{4}Cl$ (molar mass $53.5gm$ ${mol}^{-1}$) when dissolved in $1000gm$ water lowered the freezing point by ${0.637}^{o}C$. The degree of hydrolysis of the salt is $x\times {10}^{-6}$, if its degree of dissociation is $0.75$. The molal depression constant of water is $1.86kg{mol}^{-1}K$. Identify $x/9$ (nearest integer).

Answer 1

The molar concentration is the ratio of the number of moles to the volume of the solution. The number of moles is the ratio of the mass to the molar mass.
Hence, the molar concentration is $C=\frac{\frac{10}{53.5}}{1}=0.1869M$
The relationship between the hydrolysis constant, the ionic product of water and the base dissociation constant is
$K_h=\frac{K_w}{K_b}=\frac{1\times {10}^{-14}}{1.8\times {10}^{-5}}=5.5\times {10}^{-10}$
The degree of the hydrolysis is
$h=\sqrt{\frac{K_h}{C}}=\sqrt{\frac{5.5\times {10}^{-10}}{0.189}}=54\times {10}^{-5}=x\times {10}^{-5}$
Hence $x=54$