1.

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

Answer»

0.45 g
0.90 g
1.08 g
1.26 g

Solution :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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