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Question
Calculate the depression in the freezing point of water when 10 g of CH3CH2CHClCOOH is added to 250 g of water. Ka = 1.4 × 10−3, Kf = 1.86 K kg mol−1.
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Solution
Molar mass of CH3CH2CHClCOOH = 122.5 g mol−1
∴ No. of moles present in 10 g of CH3CH2CHClCOOH = `10/122.5` mol
= 8.16 × 10−2 mol
Molality of the solution (m) = `(8.16 xx 10^-2)/250 xx 1000` = 0.3264
Let α be the degree of dissociation of CH3CH2CHClCOOH
CH3CH2CHClCOOH undergoes dissociation according to the following equation:
| \[\ce{CH3CH2CHClCOOH -> CH3CH2CHClCOO- + H+}\] | |
| Initial conc. | (C mol L−1) 0 0 |
| At equilibrium | C(1 − α) Cα Cα |
∴ Kα = `(C alpha * C alpha)/(C(1 - α)) ≃ Cα^2`
or, α = `sqrt (K_α//C) = sqrt ((1.4 xx 10^-3)/0.3264)` = 0.065
To calculate the van’t Hoff factor:
| \[\ce{CH3CH2CHClCOOH -> CH3CH2CHClCOO- + H+}\] | |
| Initial conc. | 1 0 0 |
| At equilibrium | (1 − α) α α |
Total moles = 1 + α
i = 1 + 0.065
∴ i = 1.065
ΔTf = i Kf m
= 1.065 × 1.86 × 0.3264
= 0.649 ≈ 0.65°C
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