$3 \times 10^{-3} \ kg$ acetic acid is added into $500 \ cm^{3}$ water. If dissociation of acetic acid is $23\%$ then find out depression in freezing point? $K_f$ of water $= 1.86 \ K \ kg \ mol^{-1}$ and density $= 0.997 \ g \ cm^{-3}$.

$1$. Calculate moles of acetic acid $(CH_3COOH)$: Molar mass $= 60 \ g \ mol^{-1}$. Mass $= 3 \times 10^{-3} \ kg = 3 \ g$. Moles $(n) = \frac{3 \ g}{60 \ g \ mol^{-1}} = 0.05 \ mol$.
$2$. Calculate mass of solvent (water): Volume $= 500 \ cm^3$,Density $= 0.997 \ g \ cm^{-3}$. Mass $= 500 \times 0.997 = 498.5 \ g = 0.4985 \ kg$.
$3$. Calculate molality $(m)$: $m = \frac{0.05 \ mol}{0.4985 \ kg} \approx 0.1003 \ mol \ kg^{-1}$.
$4$. Calculate Van't Hoff factor $(i)$: For $CH_3COOH \rightleftharpoons CH_3COO^- + H^+$,$i = 1 \alpha = 1 0.23 = 1.23$.
$5$. Calculate depression in freezing point $(\Delta T_f)$: $\Delta T_f = i \times K_f \times m = 1.23 \times 1.86 \times 0.1003 \approx 0.229 \ K$.