An aqueous solution freezes at – 0.1860°C (${\mathrm{K}}_{\mathrm{f}}$ = 1.860; ${\mathrm{K}}_{\mathrm{b}}$ = 0.5120). What is the elevation on the boiling point? 

Depression in freezing point is given by $\Delta {\mathrm{T}}_{\mathrm{f}}=-{\mathrm{K}}_{\mathrm{f}}\times \mathrm{m}$, where ${\mathrm{K}}_{\mathrm{f}}$ is the cryoscopic constant of the solution and m is the molality.

Similarly, the elevation in boiling point is given by  $\Delta {\mathrm{T}}_{\mathrm{b}}={\mathrm{K}}_{\mathrm{b}}\times \mathrm{m}$, where ${\mathrm{K}}_{\mathrm{b}}$ is the ebullioscopic constant of the solution and m is the molality.
 

Taking the molality as a constant for the solution, we get
$-\frac{\Delta {\mathrm{T}}_{\mathrm{f}}}{{\mathrm{K}}_{\mathrm{f}}}=\frac{\Delta {\mathrm{T}}_{\mathrm{b}}}{{\mathrm{K}}_{\mathrm{b}}}⟹\frac{0.186}{1.86}=\frac{\mathrm{x}}{0.512}$

Therefore, x = 0.0512

Hence, the elevation in boiling point is $0.0512°\mathrm{C}$

And the correct answer is Option D