∆Tf =i *kf *m
Weight of water = 500 * 0.997 = 498.5 g (weight = volume * denisty)
No. of moles of acetic acid
= Wt. of CH3 COOH in g/Mol. wt. of CH3 COOH = 3 *10-3 *103/60 = 0.05
Since 498.5 g of water has 0.05 moles of CH3COOH
1000 g of water has = 0.05 *1000/498.5 = 0.1
Therefore molality of the solution = 0.1
Determination of van’t Hoff factor, i
CH3COOH → CH3COO- + H+
No. of moles at start 1 0 0
No. of moles at equb. 1 – 0.23 0.23 0.23
Therefore vant Hoff factor
= No. of particles before dissociation/No. of particles after dissociation
= 1 – 0.23 + 0.23 + 0.23/1 = 1.23
Now we know that
∆Tf = i * kf *m = 1.23 = 1.86 * 0.1 = 0.2228K
ALTERNATIVE SOLUTION :
Density of water = 0.997 g/cm3
Weight of water (W) = 500 * 0.997 = 498.5 g
Weight of acetic acid (w) = 3.0 * 10-3 kg = 33.0 kg
∆Tf = 1000 * Kf *w/m * W
Given that Kf for water = 1.86 K kg-1 mol-1
mol. wt. of CH3COOH(m) = 60
(∆Tf)cal = 1000 *1.86 * 3.0/60 * 498.5 = 0.186
(Because CH3COOH is an electrolyte and 23% dissociated)
CH3COOH ⇌ CH3COO- + H+
At t = 0 1 mole 0 0
At equilibrium (1 - ∝) mole ∝ gm-ion ∝ gm-ion
No. of particles after dissociation = 1 - ∝ + ∝ + ∝ = 1 + ∝
∝ for CH3COOH = 23/100 = 0.23 (on 23% dissociation)
So, no. of particles after dissociation = 1 + 0.23 = 1.23
By van’t Hoff factor
(∆Tf)obs/(∆Tf)cal = No. of particles after dissociation/No. of particles before dissociation
(∆Tf)obs/(∆Tf)cal = 1.23/1
(∆Tf)obs = 1.23 * 0.186 = 0.228 K
Hence depression in freezing point (∆Tf) = 0.228K
