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0.116 g of C(4)H(4)O(4) (A) in neutralis...

`0.116 g` of `C_(4)H_(4)O_(4) (A)` in neutralised by `0.074 g` of `Ca(OH)_(2)`. Hence protonic hydrogen `(H^(o+))` in `(A)` will be

A

1

B

2

C

3

D

4

Text Solution

Verified by Experts

The correct Answer is:
B
`(0.116 g "of" A)/(116 g (Mw A)) = (0.074 "of" Ca(OH)_(2))/(W[Ca(OH)_(2)])`
`W = 74 g -= 1 "mol" Ca(OH)_(2) -= 2 "mol" (overset(ɵ)(OH)) -= 2 "mol" (H^(o+)`
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