The freezing point of a 0.05 m BaCl_2 solut | vedclass

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Question

(A) For $BaCl_2$,the dissociation reaction is: $BaCl_2 \rightarrow Ba^{2+} + 2Cl^-$.Since the salt undergoes $100 \%$ ionisation,the van't Hoff factor

Vedclass · Accepted Answer

$-0.279$

Vedclass · Answer

(A) For $BaCl_2$,the dissociation reaction is: $BaCl_2 ightarrow Ba^{2+} + 2Cl^-$.Since the salt undergoes $100 \%$ ionisation,the van't Hoff factor $(i)$ is $1 + 2 = 3$.The depression in freezing point is calculated using the formula: $\Delta T_f = i 	imes K_f 	imes m$.Substituting the values: $\Delta T_f = 3 	imes 1.86 \ K \ kg \ mol^{-1} 	imes 0.05 \ mol \ kg^{-1} = 0.279 \ K$.The freezing point of the solution $(T_f)$ is given by: $T_f = T_f^0 - \Delta T_f = 0 \ ^oC - 0.279 \ ^oC = -0.279 \ ^oC$.

The freezing point of a $0.05 \ m \ BaCl_2$ solution in water ($100 \%$ ionisation) is about $(K_f = 1.86 \ K \ kg \ mol^{-1})$. (in $^oC$)

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