The freezing point of solution containing `0.2 g` of acetic acid in `20.0 g` of benzene is lowered by `0.45^(@)C`. Calculate the degree of association of acetic acid in benzene.
`(K_(f)=5.12 K^(@) mol^(-1) kg^(-1))`

Options

A. `9.45%`
B. `94.5%`
C. `47.2%`
D. `4.72%`

Answer

B

Explanation

` Delta T _ f = ( K _ f xx W xx 1000) /( w xx M ) `
` 0.45 = ( 5.12 xx 0.2 xx 1000)/( 20 xx M ) `
` therefore M = ( 5.12 xx 0.2 xx 1000 )/( 20 xx 0.45 ) `
` therefore M = 113.77 `
i=` ("calculated molar mass")/("observed molar mass") `
` = (60 ) /( 113.77) = 0.527 `
` alpha = ( 1 - i ) ( n ) /( ( n - 1 )) `
[since acetic acid dimerises in benzene ]
` = ( 1 - 0.527 ) ( 2 ) /(( 2 - 1 ) ) `
` alpha = 0.946 `
` therefore % ` degree of association = 94.6 %