1.

The measured freezing point depresion for a 0.1 m aqueous CH_(3)COOH solution is 0.19 ""^(@C). The acid dissociation constant, K_(a) at this concentration will be (Given K_f, the molal cryoscopic constant = 1.86 K kg mol^(-1))

Answer»

`4.76 xx 10^(-5)`
`4 xx 10^(-5)`
`8 xx 10^(-5)`
`2 xx 10^(-5)`

Solution :`Delta T_(F) = i xx K_(f) xx m`
` i = (0.19)/(1.86 xx 0.1) = 1.02`
`alpha = (i-1)/(n-1) = (0.02)/(1) = 2 xx 10^(-2)`
`K_(a) = C alpha^(2) =1 xx 10^(1-) xx(2 xx 10^(-2))^(2) = 4 xx 10^(-5)`


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