Some Basic Concepts of Chemistry - Result Question 38
####38. The percentage weight of $Zn$ in white vitriol $[ZnSO_4 \cdot 7 H_2 O]$ is approximately equal to ( $Zn=65, S=32, O=16$ and $H=1$ )
(a) $33.65 %$
(b) $32.56 %$
(c) $23.65 %$
(d) $22.65 %$
[1995]
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Solution:
- (d) Molecular weight of $ZnSO_4 \cdot 7 H_2 O$ $=65+32+(4 \times 16)+7(2 \times 1+16)=287$.
$\therefore$ percentage mass of zinc $(Zn)$
$=\frac{65}{287} \times 100=22.65 %$
