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Problem on Neutralization

What weight of hydrated oxalic acid should be added for complete neutralisation of 100 ml of 0.2N - NaOH solution?

(a) 0.45 g  (b)0.90 g  (c) 1.08 g  (d) 1.26 g     

Answer: (d) For complete neutralization equivalent of oxalic acid = equivalent of NaOH =
w/eq. wt. = NV/1000
w/63 = (0.2 x 100)/1000
w = 1.26 gm

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