Question

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Answer 1

36.

n-factor of Oxalic acid is 2 and of KMnO4 is 5

So just equate the no. of milliequivalents

0.1 * 2 * V = 0.025 * 20 * 5

V = 12.5 ml

35. In this one to do the question fastly you must notice that when treated with NaOH ethanoic acid will become CH3COONa or in solution it will be present as CH3COO- and when completely hydrolysed it will give ethane (C2H6).

So total moles of C from starting = total moles of C in the end

Or, Moles of ethanoic acid in the starting = moles of ethane

Moles of ethane = 0.01

Volume at STP = moles * 22400 ml = 224 ml

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