When $0.01 \ mol$ of sodium sulphate $(Na_2SO_4)$ is dissolved in $1 \ kg$ of water,complete ionization of the solution is observed. Calculate the decrease in the freezing point of the solution. $(K_f = 1.86 \ K \ kg \ mol^{-1})$
(0.0558 K) The dissociation of sodium sulphate is given by: $Na_2SO_4 \rightarrow 2Na^{+} + SO_4^{2-}$.
Since complete ionization occurs,the van't Hoff factor $(i)$ is $3$.
The molality $(m)$ of the solution is $0.01 \ mol / 1 \ kg = 0.01 \ m$.
The depression in freezing point $(\Delta T_f)$ is calculated using the formula: $\Delta T_f = i \times K_f \times m$.
Substituting the values: $\Delta T_f = 3 \times 1.86 \ K \ kg \ mol^{-1} \times 0.01 \ m = 0.0558 \ K$.
Since complete ionization occurs,the van't Hoff factor $(i)$ is $3$.
The molality $(m)$ of the solution is $0.01 \ mol / 1 \ kg = 0.01 \ m$.
The depression in freezing point $(\Delta T_f)$ is calculated using the formula: $\Delta T_f = i \times K_f \times m$.
Substituting the values: $\Delta T_f = 3 \times 1.86 \ K \ kg \ mol^{-1} \times 0.01 \ m = 0.0558 \ K$.
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