Calculate the depression in the freezing point of water when 10g of `CH_(3)CH_(2)CHClCOOH` 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)CHCICOOH = 15 +14 +13 +35.5 +12 +16 +16 +1 = 122.5 g mol^(-1)`
No. of moles present in `10g CH_(3)CH_(2)CHCICOOH = (10g)/(122.5g mol^(-1)) = 0.0816 mol`
It is given that `10g CH_(3)CH_(2)CHCICOOH` is added to `250g` of water.
Molarity of the solution `= (0.0186)/(250) xx 1000 = 0.3265 mol kg^(-1)`
Let `alpha` be the degree of dissociation of `CH_(3)CH_(2)CHCICOOH`
`CH_(3)CH_(2)CHCICOOH` undrgoes dissociation according to the following equation:
`{:(,CH_(3)CH_(2)CHCICOOHharr,CH_(3)CH_(2)CHCICOO^(-),H^(+)),("Initial conc.",C mol L^(-1),0,0),("At equilibrium",C(1-alpha),Calpha,Calpha):}`
`:. K_(a) = (Calpha.C alpha)/(C(1-alpha)) = (Calpha^(2))/((1-alpha))`
Since `alpha` is very small with respect to `1,1 -alpha ~~1`
Now, `K_(a) = (Calpha^(2))/(1) rArr K_(a) = Calpha^(2)`
`rArr alpha = sqrt((K_(a))/(C)) = sqrt((1.4 xx 10^(-3))/(0.3264)) ( :' K_(a) = 1.4 xx 10^(-3))`
`= 0.0655`
Again,
`{:(,CH_(3)CH_(2)CHCICOOHharr,CH_(3)CH_(2)CHCICOO^(-),H^(+)),("Initialmoles.",1,0,0),("At equilibrium",(1-alpha),alpha,alpha):}`
Total moles at equilibrium `= 1 - alpha +alpha +alpha = 1 +alpha`
`:. i=(1+alpha)/(1) = 1.0655`
Hence, the depression in the freezing point of water is given as:
`DeltaT_(f) = iK_(f)m`
`= 1.0655 xx 1.86 kg mol^(-1) xx 0.364 mol kg^(-1)`
`= 0.65 K`