If the degree of dissociation of aqueous solution of weak monobasic acid is determined to be 0 . 3 , then…

$\underset{1-\mathrm{\alpha }}{\mathrm{HA}}\leftrightharpoons \underset{\mathrm{\alpha }}{{\mathrm{H}}^{+}}+\underset{\mathrm{\alpha }}{{\mathrm{A}}^{-}}$

Van't Hoff factor, $\mathrm{i}=1+\mathrm{\alpha }\left(\mathrm{n}-1\right)$

                            $=1+0.3\left(2-1\right)$

                            $=1+0.3$

                            $=1.3$

$∆{\mathrm{T}}_{\mathrm{f}}$ without considering dissociation $={\mathrm{k}}_{\mathrm{f}}\times$ molality

$∆{\mathrm{T}}_{\mathrm{f}}$ without considering dissociation $={\mathrm{ik}}_{\mathrm{f}}\times$ molality

$={\mathrm{\Delta T}}_{\mathrm{f}}^{\text{'}}$

${\mathrm{\Delta T}}_{\mathrm{f}}^{\text{'}}-{\mathrm{\Delta T}}_{\mathrm{f}}=0.3$ (${\mathrm{k}}_{\mathrm{f}}\times$ molality)

$\frac{{\mathrm{\Delta T}}_{\mathrm{f}}^1-{\mathrm{\Delta T}}_{\mathrm{f}}}{{\mathrm{\Delta T}}_{\mathrm{f}}}=\frac{0.3\left({\mathrm{k}}_{\mathrm{f}}\times \mathrm{molality}\right)}{{\mathrm{k}}_{\mathrm{f}}\times \mathrm{molality}}=0.3$

$\%$ value $=30$