`2.65gm` of a diacidic base was dissolved in 500 ml of water. Twenty millilitres of this solution required 12 ml of `(N)/(6)` HCl solution. Calculate the equivalent mass and molucular mass of the base.

Explanation

`12mL (N)/(6)HCI` solution `-=20mL` of base solution
or `2mL 1N HCI` solution `-= 20mL` of base solution
or `(2)/(20) xx 500mL 1N HCI` solution `-=500mL` of base solution
or `50mL 1N HCI` solution `-= 2.65g` of base
`1000 mL 1N HCI` solution `-= (2.65)/(50) xx 1000 = 53g` base Eq. mass of the base = 53
Mol. mass of the base =Eq. mass `xx` Acidity `= 53 xx 2 = 106`