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The percentage weight of Zn in which vitriol `[ZnSO_(4)*7H_(2)O]` is approximately equal to (at. Mass of Zn =65 , S= 32, O=16 and H=1)

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The percentage weight of Zn in which vitriol `[ZnSO_(4)*7H_(2)O]` is approximately equal to (at. Mass of Zn =65 , S= 32, O=16 and H=1)
A. `33. 65%`
B. `32. 56%`
C. `23.65%`
D. `22.65%`
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Correct Answer - D
Molecules weight of
`ZnSO_(4)*7H_(2)O=65+32+(4xx16)+7(18)=287`
`:.` Percentage weight of Zn`=(65)/(287)xx100 =22 . 65%`
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