Acetic acid dimerises in benzene due to hydrogen bonding.One mole of acetic acid is added to 250g of benzene.k_b of benzene is 2kgmole^-1.The boiling point has increased by 6.4k,then the % dimerisation of acetic acid is?

we have
2CH3COOH-------(CH3COOH)2
initial mole of acetic acid=1
let d be the degree of dimerisation
[CH3COOH]=1-d
[CH3COOH]2 =d/2
number of mole after dimerisation =1-d/2
nuber of particle before dimerisation=1
i,vant hoff factor= 1+d/1 =1-d/2
we know that
delta Tb=iKbm
where delta Tb is the elevation in boiling point
i is the vant hoff facotr
m is the molality.here molality= 1/.250 =4
thus, 6.4=(1-d/2)4* 2
.8=1-d/2
d=.4
or the percentage of dimerisation is 40%