An aqueous solution freezes at – 0.186° C (K_f = 1.860; K_b = 0.512°). What is the elevation on boiling point?

• Depression in freezing point is given by ΔT_f​=−K_f ​× m, where 'K_f​' is the cryoscopic constant and 'm' is the molality of the solution.
• Elevation in boiling point is given by ΔT_b​=K_b​ × m, where 'K_b'​ is the ebullioscopic constant and 'm' is the molality of the solution.
   Taking molality as a constant for the solution, we have

$-\frac{∆{\mathrm{T}}_{\mathrm{f}}}{{\mathrm{K}}_{\mathrm{f}}}=\frac{∆{\mathrm{T}}_{\mathrm{b}}}{{\mathrm{K}}_{\mathrm{b}}}$

$\frac{0.186}{1.86} =\frac{\mathrm{x}}{0.512}$

Therefore x =0.0512

Hence, the elevation in boiling point is 0.0512^oC.