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  4. The percentage weight of Zn in white vitriol [ZnSO4 . 7H2O] is approximately equal to (at. mass of Zn= 65, S=32, 0=16 and H= 1)

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Q. The percentage weight of Zn in white vitriol $[ZnSO_4 . 7H_2O]$ is approximately equal to $(at. \, mass\, of \, Zn= 65, S=32, 0=16 \, and \, H= 1)$

AIPMTAIPMT 1995Organic Chemistry – Some Basic Principles and Techniques

Solution:

9. (d) Molecular weight of
$ZnSO_4 . 7H_2 O = 65 + 32 + (4 \times 46) + 7 (18)$
$\, \, \, \, \, \, \, \, \, = 287$
$\therefore $ Percentage weight of $Zn = \frac {65}{287} \times 100$
$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, 22.65%$