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)`.

Answer

`DeltaT_(f)=0.65^(@)C`

Explanation

Molecular mass of `CH_(3)CH_(2)CHClCOOH`
`=15+14+13+35.5+45=122.5g mol^(-1)`
`m=(W_(2)xx1000)/(Mw_(2)xxW_(1))=(10xx1000)/(122.5xx250)=0.3264`
If `alpha` is the degree of dissociation of the acid, then `{:(,CH_(3)CH_(2)CHClCOOH+,H_(2)O,hArr,CH_(3)CH_(2)CHClCOO^(c-),+,H_(2)O^(c-)),(,"Initialconc"Cmol L^(-1),,, 0,,0),(,"Conc at eq" C(1-alpha),,,C alpha,,2):}`
`:.K_(a)=(Calpha.Calpha)/(C(1-alpha))~~Calpha^(2)or alpha=sqrt(K_(a)//C)=sqrt((1.4xx10^(-3))/(0.3264))`
`=0.065`
Van't Hoff factor is `:`
`{:(,CHM_(3)CHM_(2)CHClCOOH+,,hArr,CH_(3)CH_(2)CHClCOO^(c-),+,H^(o+)),(,"Initial moles" 1,,, 0,,0),(,"Mol es of eq" (1-alpha),,, alpha,,alpha):}`
Total moles at equilibrium `=1-alpha+alpha+alpha=1+alpha`
`i=(1+alpha)/(1)=1+0.065=1.065`
`DeltaT_(f)=iK_(f)m=1.065xx1.86xx0.3264`
`=0.65^(@)C`