19.5g of `CH_(2)FCOOH` is dissolved in 500g of water. The depression in 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 of acid :
`DeltaT_("f")=1^(@)C`
`K_("f")=1.86" K kg"//"mole"^(-1)`
`DeltaT_("f")="i K_("f")"m"`
`m=(19.5)/(78)xx(1000)/(500)=0.5" m"`
`i=(1)/(1.86xx0.5)=1.0753`
Calculation of degree of dissociation of the acid :
`"CH"_(2)"F COOH"iff"CH"_(2)"F COO"^(-)+"H"^(+)`
`{:("Initial no. of moles",,,1,,,,,0,,,,,0),("Moles at equilibrium",,,1-alpha,,,,,alpha,,,,,alpha):}`
Total `=C(1+alpha)`
`i=(C(1+alpha))/(C)=1+alpha`
`alpha=i-1=1.0753-1=0.0753`
Calculation of dissociation constant for the acid :
`C=0.5" m"`
`K_(a)=([CH_(2)FCOO^(-)][H^(+)])/([CH_(2)"F COOH"])=(Calpha.Calpha)/(C(1-alpha))=(Calpha^(2))/(1-alpha)`
`K_(a)=(0.5(0.0753)^(2))/(1-0.0753)=(0.5(0.0753)^(2))/(0.9247)=3.07xx10^(-3)`.