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The percentage weight of Zn in white vit...

The percentage weight of Zn in white 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%`

Text Solution

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The correct Answer is:
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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