Solutions Part 1

As $∆{\mathit{T}}_{\mathit{b}}{}^{\mathit{b}}={\mathit{T}}_{\mathit{b}}-{\mathit{T}}_{\mathit{b}}{}^{\mathit{o}}$
Where ${\mathit{T}}_{\mathit{b}}=\mathit{b}$. pt of solution
${\mathit{T}}_{\mathit{b}}^{\mathit{o}}=\mathit{b}.\mathit{p}\mathit{t}$ of solvent or
${\mathit{T}}_{\mathit{b}}={\mathit{T}}_{\mathit{b}}^{\mathit{o}}+∆{\mathit{T}}_{\mathit{b}}$
NOTE : Elevation in boiling point is a colligative property, which depends upon the no. of particles. Thus greater the number of particles, greater is it elevation and hence greater will be its boiling point.
${\mathrm{Na}}_2{\mathrm{SO}}_4\rightleftharpoons 2{\mathrm{Na}}^{+}+{\mathrm{SO}}_4^{2-}$
Since ${\mathrm{Na}}_2{\mathrm{SO}}_4$ has maximum number of particles (3) hence has maximum boiling point.