Question

The percentage weight of Zn in white vitriol [ZnSO₄.7H₂O] 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 %

Answer 1

total weight of compound

65+32+4 (16)+7(2)(1)+7 (16)

65+32+64+14+112

=287

percent of zn

65/287*100

=22.648

correct option is d.22.65%

Answer 2

Answer: The correct answer is Option d.

Explanation:

We are given:

Mass of zinc in compound = 65 g

Mass of vitriol =  $[65+32+(4\times 16)+7((2\times 1)+16)]=287g$

To calculate the mass percent of an element in a compound, we use the equation:

$\text{Mass percent of zinc}=\frac{\text{Mass of zinc in compound}}{\text{Mass of vitriol}}\times 100$

Putting values in above equation, we get:

$\text{Mass percent of zinc}=\frac{65g}{287g}\times 100=22.65\%$

Hence, the mass percentage of zinc in the vitriol is 22.65 %.