Question

Q10.
Three moles of a solute A associate in benzene to form A. Calculate the freezing point of its
0.25 m solution if degree of association of A is 0.8 freezing point of pure benzene is 5.5°C
and K, = 5.13 k/m.​

Answer 1

Answer:

4.90 C

Explanation:

Three particles of a solute A associate to form Benzene =  (A)3  (Given)

Volume of solution = 0.25 m (Given)

Degree of association of A = 0.8 (Given)

Thus, n = 3 ∝ =0.8

∝ = i-1/1/n-1

0.8 = i-1/1/3-1

i = 0.467

Depression in freezing point ∆ Tf =i x Kf x m

∆ Tf = 0.467 x5.12 x 0.25 = 0.6 c

Freezing point of solution Tf =Tf0-∆ Tf

= 5.5-0.6

= 4.90 C