M=d×V=1.02×1.2=1.224gm
Moles of acetic acid =0.0204 moles in 2L
So molality =0.0102mol/kg
Now ΔTf=i×Kf×m
i=1+α for acetic acid
0.0198=(1+α)×1.85×0.0102
α=0.04928
≅5
1.2mL of acetic acid is dissolved in water to make 2.0L of solution. The depression in freezing point observed for this strength of acid is 0.0198∘C. The percentage of dissociation of the acid is (Nearest integer) [Given : Density of acetic acid is 1.02gmL−1 Molar mass of acetic acid is 60gmol−1
Kf(H2O)=1.85Kkgmol−1]
Held on 29 Jun 2022 · Verified 6 Jul 2026.
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