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To 500cm^(3) of water, 3.0 xx 10^(-3) kg...

To `500cm^(3)` of water, `3.0 xx 10^(-3) kg` acetic acid is added. If `23%` of acetic acid is dissociated, what will be the depression in freezing point? `K_(f)` and density of water are `1.86 K kg mol^(-1)` and `0.997 g cm^(-3)` respectively.

Text Solution

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The correct Answer is:
`DeltaT_(f)=0.228^(@)C`
Density of water=`0.997 g cm^(-3)`
Weight of water `(W_(1)) =500xx 0.997=498.5 g`
Weight of acetic acid `(W_(2))=3.0 xx 10^(-3) kg =3.0 g`
`DeltaT_(f)=(1000xxK_(f)xxW_(2))/(Mw_(2)xxW_(1))`
(Given that `K_(f)` for water =`1.86 K kg^(-1)`
Molecular weight of `CH_(3)COOH (Mw_(2))=60`
`(DeltaT_(f))_(cal)=(1000xx1.86xx3.0)/(60xx498.5)=0.186` (Because `CH_(3)COOH` is an electrolyte and `23%` dissociated)
`CH_3COOHhArrCH_3COO^(ө)+H^(o+)`
`{:(At t=0,1mol,0,0),(At Eq,(1-alpha)mol,alpha,alpha):}`
Number of particles after dissociation=`1-alpha+alpha+alpha`
=`1+alpha`
`alpha` for `CH_(3)COOH=23/100=0.23` (or 23% dissociated)`
So, number of particles after dissociation =`1+0.23=1.23`
By Van't Hoff factor
`(DeltaT_(f))_(obs)/(DeltaT_(f))_(cal)="Number of particles after dissociation"/"Number of particles before dissociation"`
`(DeltaT_(f))_(obs)/(DeltaT_(f))_(cal)=1.23/1`
`(DeltaT_(f))_(obs)=1.23 xx 0.186=0.228 K`
Hence, depression in freezing point `(DeltaT_(f))=0.228 K`
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