19.5 g of CH_(2)FCOOH is dissolved in 500 g of water. The depression in the freezing point observed is 1.0^(@)C. Calculate the van't Hoff factor and dissociation constant of fluoroacetic acid. K_(f) for water is "1.86 K kg mol"^(-1).

19.5 g of CH_(2)FCOOH is dissolved in 500 g of water. The depression i

1.	19.5 g of CH_(2)FCOOH is dissolved in 500 g of water. The depression in the freezing point observed is 1.0^(@)C. Calculate the van't Hoff factor and dissociation constant of fluoroacetic acid. K_(f) for water is "1.86 K kg mol"^(-1).
Answer» Solution :Here, `w_(2)=19.5g, w_(1)=500g, K_(f)="1.86 K kg mol"^(-1),(DeltaT_(f))_("obs")=1.0^(@)` `therefore""M_(2)"(observed)"=(1000K_(f)w_(2))/(w_(1)DeltaT_(f))=(("1000 g kg"^(-1))("1.86 K kg mol"^(-1))("19.5 g"))/("(500 g)""(1.0 K)")="72.54 g mol"^(-1)` `M_(2)"(calculated) for "CH_(2)FCOOH=14+19+45="78 g mol"^(-1)` `"van't Hoff FACTOR (i)"=((M_(2))_("cal"))/((M_(2))_("obs"))=(78)/(72.54)=1.0753.` Calculation of dissociation constant. Suppose degree of dissociation at the given concentration is `ALPHA`. `{:("Then",CH_(2)FCOOH,hArr,CH_(2)FCOO^(-),+,H^(+),),("Initial","C mol L"^(-1),,,,,),("At EQM.",C(1-alpha),,Calpha,,Calpha",","Total "=C(1+alpha)):}` `therefore""i=(C(1+alpha))/(C)=1+alpha"or"alpha=i-1=1.0753-1=0.0753` `K_(a)=([CH_(2)FCOO^(-)][H^(+)])/([CH_(2)FCOOH])=(Calpha.Calpha)/(C(1-alpha))=(Calpha^(2))/(1-alpha)` Taking volume of the solution as 500 mL, `C=(19.5)/(78)xx(1)/(500)xx1000=0.5M""therefore""K_(a)=(Calpha^(2))/(1-alpha)=((0.5)(0.0753)^(2))/(1-0.0753)=3.07xx10^(-3)`.