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")=1.86" K kg mol"^(-1)`.

Explanation

Calculation of degree of dissociation :
Mass of solute = 10 gms
Molar mass of solute `(CH_(3)-CH_(2)-CH-Cl-COOH)=122.5" g"//"mole"`
Molality `=(10)/(122.5)xx(1000)/(250)=0.326" m"`
`"CH"_(3)-"CH"_(2)-"CHClCOOH"iff"CH"_(3)"CHClCOO"^(Theta)+H^(+)`
`{:("Initial conc",,,"C moles/kg",,,0,,,0),("At equilibrium",,,C(1-alpha),,,C alpha,,,C alpha):}`
`K_(a)=(Calpha+Calpha)/(C(1-alpha))=Calpha^(2)`
Assume `1-alpha=1` for dilute solution
`alpha=sqrt((K_(a))/(C))=sqrt((1.4xx10^(-3))/(0.326))=0.065`
Calculation of Van't Hoff factor :
`CH_(3)CH_(2)CHClCOOHiffCH_(3)CH_(2)CHClCOO^(-)+H^(+)`
`{:("Initial no. of moles",,,1,,,,,0,,,,,0),("Moles at equilibrium",,,1-alpha,,,,,alpha,,,,,alpha):}`
Total no. of moles after dissociation `=1-alpha+alpha+alpha=1+alpha`
Van't Hoff factor (i) `=("Total no. of moles after dissociation")/("No. of moles before dissociation")`
`=(1+alpha)/(1)=1+alpHa=1+0.065=1.065`
Calculation of depression in freezing point
`DeltaT_("f")=i xx K_("f")xx m=1.065xx1.86xx0.326=0.65" K"`