Question

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

• A. $33.65$
• B. $32.56$
• C. $23.65$
• D. $22.65$

Answer 1

Answer: (D)

$22.65$

Answer 2

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