1.2 mL acetic acid having density 1.06 g `cm^(-3)` is dissolved in 1 litre of water. The depression in freezing point observed for this concentration of acid was 0.041`""^(@)C`. The van't Hoff factor of the acid is
( `K_(f)` of water = 1.86 K kg `mol^(-1)` )

Explanation

Number of moles of acetic acid ,
`n=(1.2xx1.06)/60=0.0212`
Molality of acid solution,
`C=0.0212/2=0.0106 "mol kg"^(-1)`
`DeltaT_f` expected for this strength = `k_f xx m`
=1.86 x 0.016 = 0.0197 K
Van.t Hoff factor (i) =`"observed freezing point"/"expected freezing point" ="-0.0205"/"-0.0197"`=1.041
Acetic acid is ionised as, `CH_3COOH hArr CH_3COO^(-) + H^(+)`
If `.alpha.` is the extent of ionisation, total number of particles =`n(1+alpha)`
van.t Hoff factor = i= `(n(1+alpha))/n=1+alpha =1.041, alpha` =0.041
Proton concentration =`C alpha` = 0.0106 x 0.041
`=4.24 xx 10^(-4) "molL"^(-1)`