Question

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

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

Answer 1

The correct option is C $128$

$25$ ml of decinormal $NaOH$ solution corresponds to $25\times \frac{0.1}{1000}=0.0025$ gram equivalents.

Hence, the molecular weight of dibasic acid will be $2\times \frac{0.16}{0.0025}=128$ g/mol.