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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.
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 cone. | (C mol L-1) | 0 | 0 |
| At equilibrium | C(1 - α) | Cα | Cα |
∴ Kα = `("C"α."C"α)/("C"(1- α)) ≃ "C"α^2`
or, α = `sqrt("K"_α//"C") = sqrt((1.4 xx 10^-3)/0.3264)` = 0.065
To calculate van't Hoff factor:
| \[\ce{CH3CH2CHClCOOH -> CH3CH2CHClCOO^- + H^+}\] | |||
| Initial moles | 1 | 0 | 0 |
| Moles at equilibrium | 1 − α | α | α |
Total moles = 1 + α
i = 1 + 0.065
∴ i = 1.065
ΔTf = iKfm = 1.065 × 1.86 × 0.3264
= 0.649 ≈ 0.65°C
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