`7.35` g of a dibasic acid was dissolved in water and diluted to 250 mL. 25 mL of this solution was neutralised by 15 mL of N NaOH solution . Calculate eq. wt and mol.wt of the acid .

Explanation

Let the equivalent weight of the acid be E.
Equivalent of acid ` = (7.35)/E " " ….(Eqn.4i)`
m.e of the acid ` = (7.35)/E xx 1000 = 7350/E " " …(Eqn.3)`
Now ,
250 mL of the acid contains `7350/E xx 1000 =7350/E m.e `
` :. 25 mL ` of the acid contains `(735)/E ` m.e .
Again, m.e of 25 mL of the acid = m.e of NaOH ... (Eqn.2)
` 735/E = 1 xx 15 `
` E = 735/15 = 49`
` :. ` eq. wt of acid = 49 .
` :. ` eq. wt of acid = 49 .
` :. ` molecular weight of the acid = eq.wt `xx` basicity
` = 49 xx2 `
` = 98 `