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
Correct option: (c) 128
Given that 0.16 g of dibasic acid required 25 ml of decinormal NaOH solution for complete neutralisation.
So, 25 ml of decinormal NaOH solution = 25 × (0.1/1000) = 0.0025 gram equivalents
The molecular weight of the acid = 2 × (0.16/0.0025) = 128 g/mol

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