ACID NaOH; \[{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\] \[\FRAC{W}{E}\times 1000=\frac{1}{10}\times 25\]; \[\frac{0.16}{E}\times 1000=\frac{25}{10}\]      \[M=2\times E=2\times 64=128\].

"> 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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0.16 g of dibasic acid required 25 ml of decinormal NaOH solution for complete neutralisation. The molecular weight of the acid will be    [CPMT 1989]

Joint Entrance Exam - Main (JEE Main) Chemistry in Joint Entrance Exam - Main (JEE Main) 10 months ago

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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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