Problem on Neutralization, What weight of hydrated oxalic acid should be added for

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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