19.5 g of CH_(2)FCOOH is dissolved in 500 g 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.

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 of water observed is 1.0^(@)C. Calculate the van't Hoff factor and dissociation constant of fluoroacetic acid.
Answer» Solution :`w_(1)=500 g "" w_(2)=19.5 g` `K_(f)=1.86" K kg mol"^(-1) "" Delta T_(f)=1 K` We known that : `M_(2)=(K_(f)XX w_(B)xx 1000)/(Delta K_(f)xx w_(1))` `= (1.86" K kg mol"^(-1)xx19.5 g xx 1000 g kg^(-1))/(500 g xx 1K)` `= 72.54 mol^(-1)` Therefore, observed molar mass of `CH_(2)FCOOH, (M_(2))_(obs)=72.54` g mol The calculated mass of `CH_(2)FCOOH` is : `(M_(2))_(cal)=14+19+12+16+16+1` `= 76 mol^(-1)` Therefore, van.t HOFF factor, `i=((M_(2))_(cal))/((M_(2))_(obs))` `= (78 g mol^(-1))/(72.54 g mol^(-1))=1.0753` Let ABE the degree of dissociation of `{:("At",CH_(2)FCOOH,hArr,CH_(2)FCOO^(-),+,H^(+)),("equilibrium","C mol L"^(-1),,O,,O),(,C(1-alpha),,C alpha,,C alpha):}` Total `= C(1-alpha)` `i=1+alpha` `alpha =i-1=1.0753-1` = 0.0753 Now, the value of `K_(a)` is given as : `K_(a)=([CH_(2)FCOO^(-)][H^(+)])/([CH_(2)FCOO])` `=(C alpha C alpha)/(C(1-alpha))=(C alpha^(2))/(1-alpha)` Taking the volume of the solution as 500 mL, we have the concentration : `C=((19.5)/(78))/(500)xx1000 M` =0.5 M Therefore, `K_(a)=(C alpha^(2))/(1-alpha)` `= (0.5(0.0753)^(2))/(1-0.0753)` `= (0.5xx0.00567)/(0.9247)` = 0.00307 (approximately) `= 3.07xx10^(-3)`