Question

0.6 ml of glacial acetic acid with density 1.06 gm L⁻¹ is dissolved in 1 kg water and the solution froze at -0.0205°C. Calculate van't Hoff factor and Kf for water is 1.86 K kg mol⁻¹ (i=1.041)

Answer 1

∆Tf (observed)= 0-(-0.0205)= 0.0205°C or K.

Now, W1 = 1 kg

Kf = 1.86 K•kg / mol

M2 = 60 g/mol (CH3COOH molar mass)

Density according to Google = 1.06 g/mL (not L)

W2 = density × volume

= 1.06 g/mL × 0.6 mL

= 0.636 g

∆Tf (calculation) = (Kf × W2)/(M2 × W1).

= (1.86 × 0.636)/(60×1)

= 0.019716 K

Vant Hoff Factor i = Observed ∆Tf/ Theoretical ∆Tf

= 0.0205/(0.019716)

= 1.03976

= 1.04

Now, CH3COOH dissociates partially to give CH3COO- and H+ Both being unimolar. Thus n' = 2.

Now, m (molality of CH3COOH) = no.of moles/1kg water. m = 0.636 / 60

= 0.0106 mole/kg

Let $ be the degree of dissociation.

$ = (i - 1)/(n' - 1)

= 0.04 / 1

= 0.04

And finally, Ka = (m × $^2)/(1-$)

= (0.0106 × 0.04 × 0.04)/(1 - 0.04)

= 1.766 × 10^-5

= 1.8 × 10^-5

Finis.

I'm tired.