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.

Explanation

Mass of solute `= 3.0 xx 10^(-3) kg = 3.0 g`
Molecular mass solute = 60
Mass of solvent `= 500 xx 0.997 = 498.5 g`
We know that,
Degree of dissociation `(alpha)=(i-1)/(n-1)`
`0.23=(i-1)/(2-1)`
or`i=1.23`
`DeltaT=ixxK_(f)xx` molality
`=1.23xx1.86xx(3xx1000)/(60xx498.5)`
`=2.229`
i.e., depression in freezing point = 0.229.