Relative lowering of vapour pressure is maximum for

We know that

Relative lowering of vapour pressure$={\mathrm{P}}_{\mathrm{A}}={\mathrm{ix}}_{\mathrm{A}}{\mathrm{P}}_{\mathrm{A}}^0$

where $\mathrm{i}$ = Van't Hoff Factor

         ${\mathrm{x}}_{\mathrm{A}}$ = concentration

         ${\mathrm{P}}_{\mathrm{A}}^0$= vapour pressure

As we can see concentration is same for all here, so now we have to look at $\mathrm{i}$(van’t hoff factor)

So

For $0.1 \mathrm{m}$ $\mathrm{urea}$, $\mathrm{i}=1$(non-electrolyte)

      $0.1 \mathrm{m} \mathrm{NaCl}$, $\mathrm{i}=2$

      $0.1 \mathrm{m} \mathrm{Mg}{\mathrm{Cl}}_2$, $\mathrm{i}=3$       

      $0.1 \mathrm{m} {\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3,$$\mathrm{i}=5$

 Since vapour pressure is directly proportional to$\mathrm{i}$.

Therefore, $0.1 \mathrm{m} {\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$ has maximum vapour pressure.

Hence, option(d) is correct.