12.25g of `CH_(3)CH_(2)CHClCOOH` ia added to 250g of water to make a solution. If the dissociation constant of above acid is `1.44 xx 10^(-3)`, the depression in freezing point of water in `.^(@)C` is : (`K_(f)` for water is `1.86 "K kg mol"^(-1)`)

Options

A. 1.789
B. 0.394
C. 1.183
D. 0.592

Answer

A

Explanation

`m=(w_(B)xx1000)/(m_(B)xxw_(A))`
`=(12.25xx1000)/(122.5xx250)=0.4`
`K_(a)=Calpha^(2)`
`1.44xx10^(-3)=0.4xx(alpha)^(2)`
`alpha = 0.06`
`alpha=(i-1)/(n-1)`
`0.06 = (i-1)/(2-1)" ":. i = 1.06`
`DeltaT=ixxK_(f)xxm`
`=1.06xx1.86xx0.4=0.789`].