Chapter 1: Problem 68
The percentage weight of \(\mathrm{Zn}\) in white vitriol \(\left[\mathrm{ZnSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\right]\) is approximately equal to \((\mathrm{Zn}=65\) \(\mathrm{S}=32, \mathrm{O}=16\) and \(\mathrm{H}=1\) ) (a) \(21.56 \%\) (b) \(32.58 \%\) (c) \(22.65 \%\) (d) \(26.55 \%\)
Short Answer
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.