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The percentage weight of Zn in white vitriol [ZnSO4·7H2O] is approximately equal to (Zn = 65, S = 32, O = 16 and H = 1):

1. 33.65%
2. 32.56%
3. 23.65%
4. 22.65%

Correct answer: 33.65%

Solution

The molar mass of ZnSO4·7H2O is calculated as 65 (Zn) + 32 (S) + 64 (O in SO4) + 126 (7H2O) = 287 g/mol. The percentage of Zn is (65/287) × 100 ≈ 33.65%.

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