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

Molecular mass of `CH_(2)FCO OH`
` = 2 xx 12+3xx1+1xx19+2 xx 16 = 78`
Molality ` = (0.25 xx 100)/(500) = 0.50m`
`Delta T_(f) = K_(f) xx m`
` = 1.86 xx 0.50 = 0.93 K`
Van.t Haff factor
` = ("Observed freezing point depression")/("Calculated frezzing point depreesion")`
`= (1.0)/(0.93) = 1.0753`
`CH_(2) FCO OH` dissociates as :
`CH_(2)FCOOH hArr CH_(2)FCOO^(-) + H^(+)`
Initial conc. `m" "0" "0`
After dissociation `m(1-alpha)" "m alpha" "m alpha`
(`alpha` is degree of dissociaiton).
Total number of moles ` = m(1-alpha) + m alpha + m alpha = m(1+alpha)`
`i = (m(1+alpha))/(m) = 1 + alpha = 1.0753`
`therefore " "alpha = 1.0753 - 1 = 0.0753`
`[CH_(2)FCO O^(-)] = m alpha= 0.50 xx 0.0753 = 0.03756`
`[H^(+)] = m alpha = 0.50 xx 0.0753 = 0.03765`
`[CH_(3)CO OH] = m (1-alpha)`
` = 0.50 ( 1-0.0753) = 0.462`
`K_(alpha) = ([CH_(2)FCOO^(-)][H^(+)])/([CH_(2)FCOOH])`
` = ((0.0365) xx (0.03765))/(0.462)`
` = 3.07 xx 10^(-3)`