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What weight of hydrated oxalic acid should be added for complete neutralisation of 100 ml of 0.2N - NaOH solutionA. 0.45 gB. 0.90 gC. 1.08 gD. 1.26 g |
Answer» Correct Answer - D For complete neutralization equivalent of axalic acid = equivalent of NaOH = `(w)/(eq.wt)=(NV)/(1000) therefore (w)/(63)=(0.2xx100)/(1000)rArr w=1.26 gm`. |
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