The freezing point of benzene decreases by `0.45^(@)C` when `0.2 g` of acetic acid is added to `20 g` of benzene. IF acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be
`(K_(f) "for benzene" = 5.12 K kg mol^(-1))`

Options

A. 0.746
B. 0.946
C. 0.646
D. 0.804

Answer

b

Explanation

`"Molality (m)"=("No.of moles of "CH_(3)COOH)/("Mass of benzene in kg")`
`((0.2))/((60 g mol^(-1))xx(0.02kg))`
`=0.167 mol kg^(-1)=0.167m`
`DeltaT_(f)=ixxK_(f)xxm or i=(DeltaT_(f))/(K_(f)xxm)`
`i=((0.45K))/((5.12 "K kg mol"^(-1))xx(0.167"mol kg "^(-1)))=0.526`
`alpha=((i-1))/((n-1))=((0-526-1))/((0.5-1))`
`=((-0474))/((-0.5))=0.948 %`