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)`)

Options

A. `4.76 xx 10^(-5)`
B. `4 xx 10^(-5)`
C. `8 xx 10^(-5)`
D. `2 xx 10^(-5)`

Answer

B

Explanation

`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)`