9.5 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 fluroacetic acid.

Explanation

Given `w_2 = 19.5 , w_1 = 500 g, K_f = 1.86 K kg "mol"^(-1) ,(Delta T_f)_(obs) = 1.0^@C`
` therefore M_2 `(observed ) `= (1000K_f w_2)/(w_1 Delta T_f) = (1000 xx 1.86 xx 19.5)/(500 xx 1.0) = 72.54 g "mol"^(-1)`
`M_2`(calculated) for `CH_2FCOOH = 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`

` therefore i = (C(1+alpha))/(C ) =1 + alpha` or ` alpha = i-1 = 1.0753 -1=0.0753`
Dissociation constant of the acid is calculated as under :
`K_alpha = ([K_2FCOO^(-)][H^+])/([CH_2FCOOH]) =(Calpha.Calpha)/(C(1-alpha)) = (C.alpha^2)/(1- alpha)`
Concentration `C = (19.5)/(78) xx 1/500 xx 1000 = 0.5M`
` therefore K_alpha = (C alpha^2)/(1- alpha) = ((0.5)(0.0753)^2)/(1-0.0753) = 3.07 xx 10^(-3)`