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प्रश्न
When 0.4g of oxalic acid is dissolved in 40g of benzene, the freezing point of the solution is lowered by 0.45 K. Calculate the degree of association of acetic acid. Acetic acid forms dimer when dissolved in benzene.
(Kf for benzene = 5.12 K kg mol-1, at. wt. C = 12, H = 1, O = 16)
उत्तर
∆Tf = 0.45 K
i = ?
Kf = 5.12 K kg mol-1
MB of CH3COOH = 60
WA = 40 g
WB = 0.4 g
We know that
`triangle "T"_"f" = "i""K"_"f" . "W"_"B"/"M"_"B" xx 1000/"W"_"A"`
`0.45 = "i" xx 5.12 xx 0.4/60 xx 1000/40`
i = `(0.45 xx 60 xx 40)/(0.4 xx 1000 xx 5.12)`
`= (45 xx 60 xx 40)/(40 xx 1000 xx 5.12)`
`= 2700/5120 = 135/256 = 0.5273`
Let degree of association of acetic acid = α
∴ α = `("i" - 1)/(1/"n" - 1)` For CH3COOH , n = 2
`= (0.5273-1)/(1/2 -1) = 0.4727/0.5000 = 0.9514`
∴ Degree of association of acetic acid (α) = 0.9514 or 95.14%
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