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

Answer

D

Explanation

0.16 g of dibasic acid is neutralised by
= 25 m L of 0.1 M NaOH
`H_(2)X + 2NaOH = Na_(2)X + H_(2)O`
Millimoles of acis `= (1)/(2)xx`millimoles of NaOH
`= (1)/(2)xx25xx0.1 = 1.25`
Now, millimoles `= (W_(mg))/("Mol. mass")`
or Mol mass `= (W_(mg))/(millimoles) = (0.16xx1000)/(1.25) = 128`