12.25 g of `CH_(3)CH_(2)CHCICOOH` is added to 250 g of water to make a solution . If the dissociation constant of above acid is `1.44xx10^(-3)` the depression in freezing point of water in `""^(@)C "is" (K_(f)` for water is 1.86 K kg `"mol"^(-1)) `

Options

A. 0.789
B. 0.394
C. 1.183
D. 0.592

Answer

A

Explanation

Step I Calculation of degree of dissociation Molar concentration of solution
`= ("Mass of solute ")/("Molar mass of solute" xx "Mass of solvent ")xx 1000`
`= (12.25 g xx 1000)/(122.5 g mol^(-1)xx 250g)`
`[because ` Molecular weight of solute ( `CH_(3)CH_(2)CHCICOOH) = 122.5 `g/mol]
= 0.40 m
If `alpha` is the degree of dissociation of `CH_(3)CH_(3)CHCICOOH` then
`CH_(3)CH_(2) rarr CH_(3)CH_(2)CHCICOO^(-)+H^(+)`
Initial concentration `" "` C mol/kg `" "` 0`" "0`
Concentration `" " C(1-alpha)" "` Calpha`" " Calpha`
equilibrium
`:. K_(a)= (Calpha^(2))/((1-alpha))~~ Calpha^(2)`
[ considering (1-`alpha) ~~ 1` dilute solution]
`implies alpha=sqrt((K_(a))/(C))`
`implies alpha= sqrt((1.44xx10^(-3))/(0.4))=sqrt(36xx10^(-4))= 0.06`
Step II Calculation of van t - Hoff factor
`CH_(3)CH_(2)CHCICOOH rarr CH_(3)CH_(2)CHCICOO^(-)+ H^(+)`
Initial moles 1 `" " 0 " " 0`
Conc. equilibrium at `1-alpha " " alpha " alpha`
Total number of moles after dissociation
`= 1-alpha+alpha+alpha=1+alpha`
`:. van t- Hoff factor `
`(i) = ("Total number of moles after dissociation ")/("Number of moles before dissociation ")`
`i= ((1+alpha))/(1) = 1 +alpha`
`:. i=1+0.06 = 1.06`
Step III Calculation of depression in freezing point `(Delta)`.
`DeltaT_(f) = iK_(f) m = (1.06)xx 0.40xx 1.86 `
`:. DeltaT_(f) = 0.789 K`