(1)

Elevation in boiling point, $\Delta T_b=m\times i\times K_b$

or      $\Delta T_b\propto m\times i$                           (As, $K_b$ is constant)             

or      $\Delta T_b\propto \text{concentration}\times \text{no. of ions}$

For 0.01 M ${\mathrm{Na}}_2{\mathrm{SO}}_4$, $\Delta T_b=0.01\times 3K_b=0.03K_b$

For 0.015 M ${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$, $\Delta T_b=0.015\times 1\times K_b=0.015K_b$

For 0.01 M urea, $\Delta T_b=0.01\times 1\times K_b=0.01K_b$

For 0.01 M ${\mathrm{KNO}}_3$, $\Delta T_b=0.01\times 2$ $K_b=0.02K_b$

Thus, 0.01 M ${\mathrm{Na}}_2{\mathrm{SO}}_4$ solution will exhibit highest boiling point.

(1)

$i\times M\downarrow \Rightarrow \Delta T_b\downarrow$

(4)

Being a strong electrolyte, $Ba{\left(OH\right)}_2$ undergoes 100% dissociation in a dilute aqueous solution,

             ${\text{Ba(OH)}}_{2\left(aq\right)} ⟶ {\text{Ba}}_{\left(aq\right)}^{2+}+2{\text{OH}}_{\left(aq\right)}^{-}$

Thus, van’t Hoff factor $\mathrm{i}=3$

(3)

$\Delta T_b=i{K}_{b}m$

For equimolal solutions, elevation in boiling point will be higher if solution undergoes dissociation, i.e., $i>1$.

(3)

$\Delta T_f=i\times K_f\times m$

So, $\Delta T_f\propto i$ (van’t Hoff factor)

Hence, $i$ is maximum, i.e., 5 for ${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3$