An aqueous solution of a substance X boils at 100.512oC. What is the freezing point of the solution? (Kf = 1.86 k/m and Kb = 0.512 K/m)

0.93oC -0.93oC 1.86oC -1.86oC

-1.86oC

Elevation in boiling point, ΔTb = 100.512 - 100 = 0.512oC m = \(\frac{ΔT_b}{K_b}\) = \(\frac{0.512}{0.512}\) = 1 m ΔTf = Kf X m = 1.86 x 1 = 1.86oC Freezing point of solution = 0 - 1.86 = -1.86oC