Calculate the depression in the freezing point of water when `10g` of `CH_(3)CH_(2)CHClCOOH` is added to `250g` water. `K_(a)=1.4xx10^(-3),K_(f)=1.86K kg mol^(-1)`.

Explanation

Molar mass of `CH_(3)CH_(2)CHClCOOH`
`=122.5 " g mol"^(-1)`
No. of moles of `CH_(3)CH_(2)CHClCOOH` present
`=(10)/(122.5)=8.16xx10^(-2)` mole.
Molality`=(8.16xx10^(-2))/(250)xx1000=0.3264 m`
Let a be the degree of dissociation of `CH_(3)CH_(2)CHClCOOH`, then and C be the initial concentratio of `CH_(3)CH_(2)CHClCOOH`

`thereforeK_(a)=(C^(2)alpha^(2))/(C(1-alpha))=(Calpha^(2))/(1-alpha)=Calpha^(2)`
{`alpha` being very very small}
`therefore K_(a)=Calpha^(2)`
`alpha=sqrt((K_(a))/(C ))=sqrt((1.4xx10^(-3))/(0.3264))=0.065`.
Calculation of van't Hoff factor :
Initially the concentration
`CH_(3)CH_(2)CHClCOOH` will be 1 mole.

`therefore i=(1+alpha)/(1)=1+alpha=1+0.065=1.065`
`therefore Delta T_(f)=i K_(f)m`
`=1.065xx1.86xx0.3264=0.65^(@)K`