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

Volume of water =` 500 cm^(3)`
Density of water ` = 0.997 g cm^(-3)`
Wt. of water ` = "Volume " xx "Density"`
` = 500 xx 0.997 = 498.5 g`
Moles of acetic acid ` = (3.0 xx 10^(-3) xx 1000)/(60)`
(mol.wt. of `CH_(3)COOH = 60`)
Molality `= ("Moles of acetic acid")/("Wt.of water") xx 1000`
` = (0.05 xx 1000)/(498.5) = 0.1003`
Acetic acid dissociates in water as :
`CH_(3)COOH hArr CH_(3)COO^(-) + H^(+)`
Initial moles `" "1" "0" "0`
If `alpha` is the degree of dissociation,
`" " 1- alpha" "alpha" "alpha`
Total moles after dissociation ,`i=("Moles after dissociation")/("Normal moles")`
` = (1+alpha)/(1)`
`alpha = 23% = 0.23`
`therefore" " i = (1+0.23)/(1) = 1.23`
Now,`" "Delta T_(f) = i k_(f) m`
` = 1.23 xx 1.86 xx 0.1003 = 0.229`
Depression in freezing point `= 0.229^(@)`