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)

Question

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)

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%`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

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%`

4 views Answered 2023-02-05 15:29