Question

The freezing point of a solution containing 5 gram of benzoic acid (M=122 g) in 35 gram of benzene is depressed by 2.94 Kelvin what is the percentage dissociation of benzoic acid if it forms a dimer in solution .
(KF for benzene = 4.9 Kelvin kg mole inverse).


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Answer 1

The percentage dissociation of benzoic acid if it forms a dimer in solution is 98%

1. We know , ∆Tf = i.Kf.m

2. Given that

∆Tf = 2.94

Kf = 4.9

3. Now, molality, m = moles of solute / kg of solvent

4. Thus, m = 5/122 / 35/1000

= 5*1000/122*35

5. Now,

∆Tf= i.Kf.m

=>2.94 = i 4.9 *5*1000/122*35

=> i = 2.94*122*35/4.9*5*1000

=> i = 0.51

Now, degree of association = i-1/(1/n) -1

Here, n = 2 as it exists as dimer

So degree of dissociation = 0.51-1/0.5-1

= 0.49/0.5

= 0.98

6. Thus, percentage of degree of association

(/dissociation) = 0.98* 100%

= 98%