Question

2g of benzoic acid (c6h5cooh) dissolved in 25g of benzene shows a depression in freezing point equal to 1.62k. Molal depression constant for benzene is 4.9 kg mol1. What is the percentage association of acid if it forms dimer in solution?

Answer 1

Given:

Quantity of benzoic acid = WA = 2 g

Quantity of Benzene =  WB= 25 g,

Molal depression =  Kf = 4.9 K kg mol–1

Freezing point =  Tf = 1.62 K

To Find:

Percentage association of acid formed.

Solution:

ΔTf = kf × Wb/Mb × 1000/Wa

= 4.9 × 2 × 1000/ 1.62 × 25

= 241.98 mol

Now,

Let the degree of association = x,

Thus, the mole of benzoic acid left undissociated = 1 - x

Therefore, total number of particles moles -

1 - x + x/2 = 1 - x/2 = i

i = Normal molecular mass/ Abnormal molecular mass

= 1- x/2 = 122/ 241.98

= 0.992

= 99.2%

Answer: The percentage of association of acid is 99.2%.

Answer 2

Answer:

Explanation:

answer is 242 g