0.115 g of a dibasic acid required 25 mlof 0.1NNaOHsolution for complete neutralisation. Calculate the molecular mass of acid

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Solution

$25 m l o f 0.1 N o f N a O H = 0.115 g 1000 m l o f 0.1 N N a O H = \frac{0.115}{25 \times 0.1} \times 1000 \times 0.1 = 4.6 g 1000 m l o f 1 N a c i d = \frac{4.6}{0.1} \times 1 = 46 g T h u s t h e e q u i v a l e n t m a s s = 46 g A c i d i s a d i b a s i c a c i d s o n = 2 M o l e c u l a r m a s s o f t h e a c i d = E q m a s s \times n = 46 \times 2 = 92 g$

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