0.16 g of dibasic acid required 25 ml of decinormal NaOH solution for complete neutralisation. The molecular weight of the acid will be

Options

A. 32
B. 64
C. 128
D. 256

Answer

C

Explanation

No. of g. equivalents of acid = g. eq. of NaOH
` = 25 xx 10^(-3) xx 0.1`
` = 2.5 xx 10^(-3)`
Equivalent `wt. xx 2.5 x 10^(-3)` = wt. Of acid
Equivalent wt. `= (0.16 g)/(2.5 xx 10^(-3)) = 64 g `
Molecular wt. = Eq. wt × basicity
` = 64 xx 2 = 128`