Question

percentage amount of oxygen in ZnSO4.7H2O​

Answer 1

Explanation:

22.65%

Molecular weight of white vitriol ZnSO

4

.7H

2

O=65+32+(4×16)+7(2×1+16)=287 g

Weight of Zn in white vitriol = 65 g

Percentage weight of Zn =

Molecularweightofwhitevitriol

WeightofZn

×100=

287

65

×100=22.65 %