0.2 molal acid HX is 20% ionised in solution Kf = 1.86 K molality–1. The freezing point of the solution is
-0.45
-0.90
-0.30
-0.53
HX → H+ + X-
n=2
i = 1 + a(n – 1)
i = 1 + 0.2(2 – 1)
i = 1 + 0.2
i = 1.2
∆Tf = i Kf × m
∆Tf = 1.2 × 1.86 × 0.2
∆Tf = 0.446
∆Tf = F.P. of solvent – F.P. of solution
0.45 = 0 – F.P. of solution
F.P. of solution = – 0.45
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