`19.5g` of `CH_(2)FCOOH` is dissolved in `500g ` of water . The depression in the freezing point of water observed is `1.0^(@)C`. Calculate the Van't Hoff factor and dissociation constant of fluoroacetic acid.

Explanation

Calculation of Van't Hoff factor (i) for the acid
`DeltaT_(f)=iK_(f)m. or i=(DeltaT_(f))/(K_(f)m)`
`DeltaT_(f)=1.0^(@)C=1.0 K, K_(f)=1.86 K. kg mol^(-1)`,
`m=W_(B)/(M_(B)xxM_(A))=((19.5g))/((78g mol^(-1))xx(0.5 kg))=0.15 mol kg^(-1)`
`i=((1.0K))/((1.86 K kg mol^(-1))xx(0.5 mol kg^(-1)))=1.0753`
Calculation of degree of dissociation for acid
`"Molar conc. of acid"=((19.5g)//(78g mol^(-1)))/((0.5 litre))("Volume of solution is considered as 500 mL or 0.5 L")`
=0.5 ,mol `L^(-1)`=0.5 M
`{:(CH_(2)(F)COOH,hArr,CH_(2)(F)COO^(-)(aq)+H^(+)(aq)),(0.5,," "0" "0),(0.5(1-alpha),,0.5alpha" "0.5alpha):} `
`K_(a)=((0.5alpha)xx(0.5alpha))/(0.5(1-alpha))=(0.5xx0.0753)^(2)/((1-0.0753))`
`(0.5xx(0.0753)^(2))/((0.9247))=3.07xx10^(-3)`.