Calculate the depression in freezing point of water when 10 g of `CH_(3)CH_(3)CH(CI)COOH` is added to 250 g of water. `K_(a)=1.4xx10^(-3), K_(f)=1.86 K kg mol^(-1)`.

Explanation

Molar mass of ` CH_(3) CH_(2)CHCICO OH`
` = 4 xx 12+7+35.5 +2 xx 16 = 122.5 g mol^(-1)`
Moles of `CH_(3)CH_(2)CHICOOH = (10)/(122.5) = 8.16 xx 10^(-2) mol`
Molality of solution ` = (8.16 xx 10^(-12) xx 1000)/(250)`
` = 0.3265m`
If `alpha` is the degree of dissociation of `CH_(3) CH_(2)CHCIC O OH`
`CH_(3)CH_(2)CHCICOOH hArr CH_(3) CH_(2)CICO O^(-) + H^(-)`
Conc . after dissociaiton .
`e(1-alpha)" c alpha" "c alpha`
`K_(alpha) = (c alpha. c alpha)/(c(1-alpha)) = c alpha^(2)`
`or " " alpha = sqrt(K_(alpha)//c) = sqrt((1.4 xx 10^(-3))/(0.3265)) = 0.065`
To calculate Van.t Hoff factor,
`CH_(3)CH_(2)CHClCOOH hArr CH_(3)CH_(2)CHCICO O^(-) + H^(+)`
At equi `" "1-alpha" alpha" "alpha`
`i = (1- alpha + alpha + alpha)/(1) = (1 + alpha)/(1) = 1 + 0.065 = 1.065`
`DeltaT_(f) = i K_(f)m = 1.065 xx 1.065 = 1.065`
`Delta T_(f) = iK_(f) m = 1.065 xx 1.86 xx 0.3265`
`= 0.647^(@)`