R-COOH dimerises in benzene upto 50% and it ionises upto 10% in water.Detrmine the ratio of Van't hoff factors in dimerisation v/s ionisation?

Here, we know that , degree of dissociation α = no. of moles associated / total no. of moles taken.

If we take 1 mole of RCOOH in benzene , we get 50 % of it dimerised so, no of moles associated = 1/2 

Hence, α = 1/2 

For dimerisation of 1 mole of RCOOH, we need to take 2 molecules of it so no. of molecules , n= 2

The reaction is :

 ................2RCOOH ⇔  (RCOOH)₂ 

Before asso. 1 mole......... 0

after asso. 1-α  ...................α /2

Now, we know that, i = 1-α + α /n

Here, n = 2

So, van't hoff factor i =  1-α + α /2= 1-1/2 + (1/2)/2 = 3/4

Now, in case of dissociation

 ................RCOOH ⇔   RCOO^-  + H⁺ 

Before asso. 1 mole......... 0....................0

after asso. 1-α  ...................α  .....  ......α 

Van't hoff's factor i = 1+ (n-1)α/1 

Here, n = 1 so, i = 1+ α

Now, we know that dissociation is 10% , so for 1 mole,  α = 1/10 

So, i = 1+ 1/10 = 11/10

hence,ratio of ,  i of dimerisation/ i of ionization = (3/4 )/ (11/10) = 15/22