Question

Solve this correctly






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Answer 1

16 ml

Acid is present .1*100=10 millimoles

First by titration with Naoh the acid remaining is 4 millimoles

As .2*30 ml is 6 milli moles therefore it neutralise 6 milli moles of acid therefore 4 millimoles is remaining

Now 4 millimoles of Koh is req to neutralise 4 millimoles of acid

Therefore 4 = .25*V

V=16 ml

Hence 16 ml Koh is req

Hope it helps

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