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

Options

A. 0.45 g
B. 0.90 g
C. 1.08 g
D. 1.26 g

Answer

D

Explanation

For complete neutralization equivalent of oxalic acid = equivalent of NaOH =
`(w)/(eq.wt)=(NV)/(1000)therefore(w)/(63)=(0.2xx1000)/(1000)impliesw=1.26g`