A solution containing 2 . 5 × 10 - 3   kg of a solute dissolved in 75 × 10 - 3   kg of…

${\mathrm{\Delta T}}_{\mathrm{b}}={\mathrm{iK}}_{\mathrm{b}}\mathrm{m}$

$$\begin{gathered} {\mathrm{\Delta T}}_{\mathrm{b}}=\mathrm{Elevation} \mathrm{of} \mathrm{boiling} \mathrm{point} \\ {\mathrm{K}}_{\mathrm{b}}= \mathrm{molal} \mathrm{boiling} \mathrm{point} \mathrm{elevation} \mathrm{constant} \\ \mathrm{m}=\mathrm{molality} \\ \mathrm{i}=\mathrm{van}\text{'}\mathrm{t} \mathrm{hoff} \mathrm{factor} \end{gathered}$$

$\mathrm{m}=\frac{2.5}{\mathrm{M}}\times \frac{1000}{75}$

$$\begin{gathered} ∆{\mathrm{T}}_{\mathrm{b}}={\mathrm{T}}_{\mathrm{b}}-{\mathrm{T}}_{\mathrm{b}}^{\mathrm{o}} \\ {\mathrm{T}}_{\mathrm{b}}=\mathrm{Boiling} \mathrm{point} \mathrm{of} \mathrm{the} \mathrm{solution} \\ {\mathrm{T}}_{\mathrm{b}}^{\mathrm{o}}=\mathrm{Boiling} \mathrm{point} \mathrm{of} \mathrm{the} \mathrm{solvent} \\ ∆{\mathrm{T}}_{\mathrm{b}}=373.535-373.15 \\ =0.385 \mathrm{K} \end{gathered}$$

${\mathrm{\Delta T}}_{\mathrm{b}}={\mathrm{iK}}_{\mathrm{b}}\mathrm{m}$

$0.385=1\times 0.52\times \frac{2.5\times 1000}{\mathrm{M}\times 75}$

The molar mass of the solute,$\mathrm{M}=45 \mathrm{g}/\mathrm{ml}$.