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(B) The dissociation of $PbCl_2$ is given by: $PbCl_{2(s)} \leftrightarrow Pb^{2+}_{(aq)} + 2Cl^-_{(aq)}$The solubility product expression is: $K_{sp}

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$-0.06$

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(B) The dissociation of $PbCl_2$ is given by: $PbCl_{2(s)} \leftrightarrow Pb^{2+}_{(aq)} + 2Cl^-_{(aq)}$The solubility product expression is: $K_{sp} = [Pb^{2+}][Cl^-]^2 = (s)(2s)^2 = 4s^3$Given $K_{sp} = 4 	imes 10^{-6}$,we have $4s^3 = 4 	imes 10^{-6}$,which gives $s = 10^{-2} \ mol \ L^{-1}$.For $PbCl_2$,the Van't Hoff factor $i = 3$ (one $Pb^{2+}$ and two $Cl^-$ ions).The depression in freezing point is calculated as: $\Delta T_f = i 	imes K_f 	imes m$.Assuming $m \approx s = 10^{-2} \ mol \ kg^{-1}$,we get $\Delta T_f = 3 	imes 2 	imes 10^{-2} = 0.06 \ K$.Since the freezing point of pure water is $0 \ ^oC$,the freezing point of the solution is $0 - 0.06 = -0.06 \ ^oC$.

$PbCl_2$ is dissolved in water to make its saturated solution. What will be the freezing point of this solution?
Given: $K_f (H_2O) = 2 \ K \ kg \ mol^{-1}$,$K_{sp} (PbCl_2) = 4 \times 10^{-6}$
(Assume molarity to be equal to molality) $..... ^oC$

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$PbCl_2$ is dissolved in water to make its saturated solution. What will be the freezing point of this solution?Given: $K_f (H_2O) = 2 \ K \ kg \ mol^{-1}$,$K_{sp} (PbCl_2) = 4 \times 10^{-6}$(Assume molarity to be equal to molality) $..... ^oC$