In comparison to a 0.01 M solution of glucos | vedclass

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(C) The depression in freezing point is a colligative property,which depends on the number of particles in the solution.Glucose is a non-electrolyte a

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(C) The depression in freezing point is a colligative property,which depends on the number of particles in the solution.Glucose is a non-electrolyte and does not dissociate in water,so its van't Hoff factor $(i)$ is $1$.$MgCl_2$ is a strong electrolyte that dissociates as $MgCl_2 \rightarrow Mg^{2+} + 2Cl^-$,producing $3$ ions per formula unit,so its van't Hoff factor $(i)$ is $3$.Since the concentration of both solutions is $0.01 \ M$,the depression in freezing point for $MgCl_2$ will be approximately $3$ times that of the glucose solution.Therefore,the correct option is $(C)$.

In comparison to a $0.01 \ M$ solution of glucose,the depression in freezing point of a $0.01 \ M \ MgCl_2$ solution is (Molecular weight of $MgCl_2 = 95$,molecular weight of glucose $= 180$).

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