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1 g of an acid C(6)H(10)O(4) is complete...

`1 g` of an acid `C_(6)H_(10)O_(4)` is completely neutralised by `0.768 g KOH`. Calculate the number of neutralizable protons in acid.

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Verified by Experts

The correct Answer is:
2
`"Meq.of"C_(6)H_(10)O_(4)="Meq.of" KOH`
`(1)/((146)/(n))xx1000=(0.768)/(56)xx100`
( where `n` is no.of neutralizable proton)
`:. n=2`
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