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Question

The percentage weight of Zn in white vitriol [ZnSO47H2O] is approximately equal to?

A
33.65%
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B
32.56%
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C
23.65%
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D
22.65%
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Solution

The correct option is D 22.65%
Molecular weight of white vitriol ZnSO4.7H2O=65+32+(4×16)+7(2×1+16)=287 g
Weight of Zn in white vitriol = 65 g
Percentage weight of Zn = WeightofZnMolecularweightofwhitevitriol×100=65287×100=22.65 %

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