In comparison to a 0.01 M solution of glucose, the depression in freezing point of a 0.01 M MgCl2 solution is _____________.

Depression in freezing point = i × m × K​f


Now, i = van’t Hoff factor, m is the molality and Kf is the freezing constant of the solvent.


Here, we see that molality is the same i.e. 0.01 M and Kf will be the same as the solvent is water.


Thus, depression in freezing point depends only on the van’t Hoff factor.


i = Total number of ions after dissociation/association/Total number of ions before dissociation/association.


i= 3 for MgCl2 as it dissociates into 3 ions


i = 1 for glucose as it does not undergo any association/dissociation.


Thus, option (iii) is the correct option.

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