19.5 g of `CH_2FCOOH` 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.

Explanation

`W_B=19.5g, W_A= 500g=0.5 kg`
`(M_B)` Molar mass of `(FCH_2COOH)=78 "g mol"^(-1)" "DeltaT_f=1^@C=1K`
`K_F` (water) = `1.86 "k Kg mol"^(-1)`
`DeltaT_f=iK_fm" "` (Depression in Freezing Point Equation)
`DeltaT_f=iK_fxxW_B/M_Bxx1/(W_A(kg))`
`i=(DeltaT_fxxM_BxxW_A(kg))/(K_f xxW_B)`
(Vant. Hoff Equation), `i=(1xx78xx0.5)/(19.5xx1.86)=1.0753`
Fluroacetic acid dissociates as follows:
`CH_2FCOOH(aq)hArrCH_2FCOO^(-1)(aq)+H^(+)(aq)`
Initial conc. C
Eqn.Conc.`calpha " "calpha`
So , `i=(C(1-alpha)+Calpha+Calpha)/(C)=1+alpha`
`1+alpha = 1.0753`
`alpha = 0.0753`
Dissociation constant `K_a=([FCH_2" "COO^(-1)[H^(+)]])/([CH_2FCOOH])`
`=(Calpha.Calpha)/(C(1-alpha))=(Calpha^2)/(1-alpha)`
Here, `C=n_B/V=W_B/M_B xx1/V`
`=(19.5)/78xx1/(0.5L)=0.5" mol L"^(-1)`
`:. K_a=(0.5(0.0753)^2)/(0.9247)=3.07xx10^(-3)`