Question

If a compound $AB$ dissociates to the extent of $75\%$ in an aqueous solution, the molality of the solution which shows a $1.82\text{K}$ rise in the boiling point of the solution is
(Given: Molal elevation constant for water $=0.52{\text{K kg mol}}^{-1}$)

Answer 1

$\begin{array}{ccccc}AB& \to & A^{+}& +& B^{-}\\ 1-\alpha & & \alpha & & \alpha \end{array}$
Given that
$\alpha =3/4$
$N=2$
$i=[1+(2-1\left)\alpha \right]$
The rise in boiling point is
$\Delta T_b=i\times K_b\times m$
$\Rightarrow 1.82=[1+(2-1\left)3/4\right]\times 0.52\times m$
$\Rightarrow m=\frac{1.82}{0.52\times \frac{7}{4}}=\frac{7.28}{3.64}=2$
$\Rightarrow m=2\text{mol/kg}$