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(C) The dissociation of sodium sulfate is given by: $Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-}$.Since it undergoes complete dissociation,the van't Hoff f

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$0.0558$

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(C) The dissociation of sodium sulfate is given by: $Na_2SO_4 ightarrow 2Na^+ + SO_4^{2-}$.Since it undergoes complete dissociation,the van't Hoff factor $(i)$ is $3$ (two $Na^+$ ions and one $SO_4^{2-}$ ion).The formula for depression in freezing point is $\Delta T_f = i 	imes K_f 	imes m$.Here,$i = 3$,$K_f = 1.86 \ K \ kg \ mol^{-1}$,and molality $(m)$ = $\frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{0.01 \ mol}{1 \ kg} = 0.01 \ mol \ kg^{-1}$.Substituting the values: $\Delta T_f = 3 	imes 1.86 	imes 0.01 = 0.0558 \ K$.

If sodium sulfate $(Na_2SO_4)$ undergoes complete dissociation into its constituent ions in an aqueous solution,what will be the depression in freezing point $(\Delta T_f)$ when $0.01 \ mol$ of sodium sulfate is dissolved in $1 \ kg$ of water (in $K$)? (Given: $K_f = 1.86 \ K \ kg \ mol^{-1}$)

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