Question

The lowering in the vapour pressure is maximum for:
(A) $0.1{\text{M urea }}$
(B) $0.1{\text{M NaCl}}$
(C) $0.1{\text{M Mg{Cl_2}}}$
(D) $0.1{\text{M }}{{\text{K}}_4}\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_6}} \right]{\text{ }}$

Answer 1

Hint:Lowering in the vapour pressure for any solution will depend on the number of solute particles it is having in its solution. For getting the no of solute particles we will concentrate on the molarity and the no of ions of the particular solute which we will get on its dissociation.

Complete step by step solution:
> Lowering in the vapour pressure for any solution will depend upon the value of the van't hoff factor which is the measure of the effect of the solute on the colligative properties like lowering in the vapour pressure. Van't hoff factor is the ratio of the actual concentration of the particle after the dissociatrion on dissolving and the concentration of the particles when calculated from its mass.
> So here if we assume there is 100% dissociation then for the calculation of the Vant’s hoff factor.
- \[{\text{ \alpha = }}\dfrac{{{\text{i - 1}}}}{{{\text{n - 1}}}}\], where ${\text{i = }}$vant hoff factor, ${\text{n = }}$no of ions after dissociation, ${\text{ \alpha = }}$degree of dissociation$ = 1$
- Now the compound $0.1{\text{M }}{{\text{K}}_4}\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_6}} \right]{\text{ }}$,
${\text{1 = }}\dfrac{{{\text{i - 1}}}}{{{\text{5 - 1}}}}$, ${\text{n = }}$5, since there will be 5 ions on dissociation.
$ \Rightarrow {\text{4 = i - 1}}$
$ \Rightarrow {i_{{{\text{K}}_4}\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_6}} \right]{\text{ }}}} = 5$
- For compounds $0.1{\text{M NaCl}}$and $0.1{\text{M Mg{Cl_2}}}$ the value of the ${\text{n = 2}}$ and ${\text{n = 3}}$ ,so calculation for van't hoff factor for the same is as follows
 ${\text{1 = }}\dfrac{{{\text{i - 1}}}}{{{\text{2 - 1}}}}$
For $0.1{\text{M NaCl}}$ van't Hoff factor will be 2 and for $0.1{\text{M Mg{Cl_2}}}$ van't Hoff factor will be 3.
- Urea does not undergo dissociation therefore its van't hoff factor will be 1.
So here we see that the van't hoff factor for the compound is highest among the other compounds given in the other options.
And as we know that the van't hoff factor is a measure of the lowering in the vapour pressure. Hence the lowering in the vapour pressure is maximum for the compound$0.1{\text{M }}{{\text{K}}_4}\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_6}} \right]{\text{ }}$in comparison of the other three option given in the problem.
Hence option (D) is the correct answer.

Note: Relative lowering in the vapour pressure in any solution is a colligative property. Colligative properties of a solution are those that depend only on the concentration of solute particles. These properties also include the properties like vapour pressure, boiling point, and freezing point of the solvent in the solution. The colligative properties related to these properties will be considered as lowering in the vapour pressure, depression in the freezing point, elevation in the boiling point and osmotic pressure.