To $500{ \mathrm{cm}}^3$ of water, $3.0\times {10}^{-3} \mathrm{kg}$ of acetic acid is added. If 23% of acetic acid is dissociated, what will be the depression in freezing point. If ${\mathrm{k}}_{\mathrm{f}}$ and density of water are $1.86{\mathrm{kkg}}^{-1}$ and $0.997{\mathrm{gcm}}^{-3}$ respectively

Number of moles of acetic acid $=\frac{3\times {10}^{-3} \mathrm{kg}}{60{\mathrm{gmol}}^{-1}}=\frac{3 \mathrm{g}}{60{\mathrm{gmol}}^{-1}}=0.05$

Mass of water $={\mathrm{m}}_1=500{ \mathrm{cm}}^3\times 0.997{\mathrm{gcm}}^{-3}=0.4985 \mathrm{kg}$

Molality of acetic acid $=\mathrm{m}=\frac{{\mathrm{n}}_2}{{ \mathrm{m}}_1}=\frac{0.05 \mathrm{mol}}{0.4985 \mathrm{kg}}=0.1003{\mathrm{molkg}}^{-1}$

Dissociation of ${\mathrm{CH}}_3\mathrm{COOH}$ :
$$\begin{gathered} {\mathrm{CH}}_3\mathrm{COOH}\rightleftharpoons {\mathrm{CH}}_3{\mathrm{COO}}^{\ominus }+{\mathrm{H}}^{\oplus } \\ \mathrm{at} \mathrm{t}=0, 1 0 0 \\ \mathrm{at} \mathrm{equ}., 1-\mathrm{\alpha } \mathrm{\alpha } \mathrm{\alpha } \end{gathered}$$

Total number of moles $=i=1-\alpha +\alpha +\alpha =1+\alpha$

as, $\Delta {\mathrm{T}}_{\mathrm{f}}={\mathrm{iK}}_{\mathrm{f}}\mathrm{m}$
$\Delta \mathrm{T}=(1+\alpha ){\mathrm{K}}_{\mathrm{f}}\mathrm{m}=(1+0.23)\left(1.86 {\mathrm{KKgmol}}^{-1}\right)\left(0.1003 \mathrm{mol} \mathrm{K}{ \mathrm{g}}^{-1}\right)=0.228 \mathrm{K}$

Hence, the correct option (B). 0.228 k