To `500 cm^(3)` of water, `3.0xx10^(-3) kg` of 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^(-1) mol_(-1)` and `0.997 g cm^(-3)`, respectively.

Answer

`0.23k`

Explanation

`DeltaT = K_(f) xx` molality `xx (1-alpha)`
For acetic acid: `{:(CH_(3)COOH hArr,CH_(3)COOH+,H^(+)),(1,0,0),(1-alpha,alpha,alpha):}`
Given, `alpha = 0.23`, Also molality
`=("mole of acetic acid")/("weight of water in kg")`
`=(3 xx 10^(-3) xx 10^(-3))/(60xx(500 xx 0.997)/(10^(3))) = 0.10`
`DeltaT = K_(f) xx` molality `(1-alpha)`
`DeltaT = 1.86 xx 0.1 xx 1.23 = 0.229`