1.26 g of a dibasic acid were dissolved in water and the solution made up to 200 mL. 20 mL of this solution were completely neutralised by 10 mL of `N/5` NaOH solution. Calculate the equivalent mass and molecular mass of the acid.

Explanation

`12ml(N)/(6)` HCl solution `-=20ml` of base solution
or `2ml1 N HCl` solution `-=20ml` of base solution
or `(2)/(20)xx500ml1 N HCl` solution `-=500ml` of base solution
or `50 ml 1 N HCl` solution `-=2.65gm` of base
`1000ml 1 N HCl` solution `-=(2.65)/(50)xx1000=53gm` bese
Equivalent mass of the base `=53`
Molecular mass of the base
`=` equivalent mass `xx` acidity
`=53xx2=106`