The measured freezing point depression for a 0.1 m aqueous `CH_(3)COOH` solution is `0.19^(@)C`. The dissociation aconstant `(K_(b))` for the acid at this concentration will be `(K_(f)=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) = 0.19 , m = 0.1 , K_(f) = 1.86 Km^(-1)`
` Delta T_(f) = k_(f) xx m or I = (Delta T_(f))/(K_(f) xx m)`
` = (0.19)/(1.86 xx 0.1)= 1.02`
If `alpha` is the degree of dissociation
`{:(," "CH_(3)COOH ,hArr,CH_(3)COO^(-), + H^(+)),(,"Initial moles",1,0,0),(,"After dissociation",1-alpha,alpha,alpha):}`
` = 1 -alpha + alpha + alpha =1 + alpha`
`i = (1+alpha)/(1) = 1.02 or alpha = 0.02`
` k_(a) = calpha^(2) = 0.1 xx (0.02)^(2) = 4 xx 10^(-5)`