a weak electrolyte AB in 5% dissociated in aqueous solution.what is the freezing point of a 0.10 molar aqueous solution of AB ? Kf =1.86 deg/molal?
α=0.05
i= 1 + (n-1)α = 1+0.05 = 1.05
ΔTf = i Kf m = 1.051.860.10 = 0.1953 K
Tf(pure solvent) − Tf(solution) = 0.195 K
Tf(solution) = 273-0.195 = 272.8 K
i= 1 + (n-1)α = 1+0.05 = 1.05
ΔTf = i Kf m = 1.051.860.10 = 0.1953 K
Tf(pure solvent) − Tf(solution) = 0.195 K
Tf(solution) = 273-0.195 = 272.8 K