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)

Question

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) (A) 33.65% (B) 32.56% (C) 23.65% (D) 22.65%

Tardigrade Admin · Accepted Answer

9. (d) Molecular weight of ZnSO4 . 7H2 O = 65 + 32 + (4 × 46) + 7 (18) = 287 ∴ Percentage weight of Zn = (65/287) × 100 22.65%

Q. The percentage weight of Zn in white vitriol $[Z n S O_{4} .7 H_{2} O]$ is approximately equal to $(a t . ma ss o f Z n = 65, S = 32, 0 = 16 an d H = 1)$

$33.65$

$32.56$

$23.65$

$22.65$

9. (d) Molecular weight of
$Z n S O_{4} .7 H_{2} O = 65 + 32 + (4 \times 46) + 7 (18)$
$= 287$
$∴$ Percentage weight of $Z n = \frac{65}{287} \times 100$
$22.65$

Q. The percentage weight of Zn in white vitriol [ZnSO4​.7H2​O] is approximately equal to (at.massofZn=65,S=32,0=16andH=1)

33.65

32.56

23.65

22.65