Question

Of the following $ 0.10{\text{ m}} $ aqueous solutions, which one will exhibit the largest freezing point depression?
(A) $ KCl $
(B) $ {C_6}{H_{12}}{O_6} $
(C) $ A{l_2}{(S{O_4})_3} $
(D) $ {K_2}S{O_4} $

Answer 1

Hint: Freezing point depression is the phenomenon that describes why adding a solute to solvent results in the lowering of the freezing point of the solvent. When a substance starts to freeze, the molecules slow down due to the decreases in temperature, and the intermolecular forces start to take over.

Complete answer:
The depressed freezing point of a solution is less than the freezing point of the pure solvent. This means that a solution must be cooled to a lower temperature than the pure solvent in order for freezing to occur.
Its formula is $ \Delta {T_f} = i{k_f}m $ , here initial temperature will be same for all compound and the value of m is also same for all the compounds but the value of (i) which is van't hoff factor and will be different for all the compounds and the depression of freezing point will solely depend upon it.
The compound which has maximum value of ‘i’
The values for different compounds are:
 $ KCl = 2 $
 $ {C_6}{H_{12}}{O_6} = 1 $
 $ A{l_2}{(S{O_4})_3} = 5 $
 $ {K_2}S{O_4} = 3 $
Now here we can see that $ A{l_2}{(S{O_4})_3} $ has the highest Van’t hoff factor which means it will show maximum depression in freezing point.
So our answer is option C.

Note:
At the freezing point of a solvent, there exists an equilibrium between the liquid state and the solid-state of the solvent. This implies that the vapor pressures of both the liquid and the solid phase are equal. Upon the addition of a solute that is non-volatile, the vapor pressure of the solution is found to be lower than the vapor pressure of the pure solvent. This causes the solid and the solution to reach equilibrium at lower temperatures.