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Normality

0.16 g of dib...

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

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B

64

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C

128

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D

256

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Solution

The correct option is C 128
Dibasic acid NaOH; $N_{1} V_{1} = N_{2} V_{2}$
$\frac{W}{E} \times 1000 = \frac{1}{10} \times 25; \frac{0.16}{E} \times 1000 = \frac{25}{10}$
$M = 2 \times E = 2 \times 64 = 128$ .

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