0.1092 g of a dibasic acid is exactly neutralized by `21cm^(3)` of 0.1N NaOH. Calculate the molecular mass of the acid.

Explanation

Mass of acid used = 0.10928 g
Basicity of the acid = 2 ( `:.` acid is dibasic)
Volume of NaOH `(V_1)` required = 21 `cm^3`
Normality of NaOH = 0.1 N
Suppose, the equivalent mass of the acid is E.
Gram equivalents of acid = Gram equivalents of base
` therefore (0.1092 )/(E ) = ( 0.1 xx 21 )/(1000)`
or ` E=(0.1092 xx 1000 )/( 0.1 xx 21 ) = 52`
Molecular mass of the given acid
= E` xx` basicity `= 52 xx 2 = 104`.