Answer:
The appropriate answer is "9.225 g".
Explanation:
Given:
Required level,
= 63 ppm
Initial concentration,
= 22 ppm
Now,
The amount of free SO₂ will be:
= [tex]Required \ level -Initial \ concentration[/tex]
= [tex]63-22[/tex]
= [tex]41 \ ppm[/tex]
The amount of free SO₂ to be added will be:
= [tex]41\times 225[/tex]
= [tex]9225 \ mg[/tex]
∵ 1000 mg = 1 g
So,
= [tex]9225\times \frac{1}{1000}[/tex]
= [tex]9.225[/tex]
Thus,
"9.225 g" should be added.
