`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

Using the relation
`underset(("observed"))(M_(2))=(100K_(f)W_(2))/(W_(1)DeltaT_(f))`
`=(1000xx1*86xx19*5)/(500xx1)=72*54" g "mol^(-1)`
`M_(2)("calculated")=12+2+19+12+2xx16+1`
`=78" g "mol^(-1)`
`thereforei=(M_(2)("calculated"))/(M_(2)("observed"))=(78)/(72*54)=1*0753`
Calculation of dissocation constant
Let `alpha` be the degree of dissociation and C be the initial concentration of `CH_(2)FCOOH`
`{:(,CH_(2)FCOOH,hArr,CH_(2)FCOO^(-),,H^(+)),("Initially",C,,0,,0),("At equilibrium",C(1-alpha),,0,,0):}`
`i=(C(1+alpha))/(C)=1+alpha`
`alpha=i-1=1*0753-1=0*0753`
`K_(a)=(CalphaCalhpa)/(C(1-alpha))=(Calpha^(2))/(1-alpha)`
`K_(a)=(0.5xx(0.753)^(2))/((1-0.753))=3.07xx10^(-3)`.