Can someone please tell me what the answer is to all this
1 answer:
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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:
=
=
=
The amount of free SO₂ to be added will be:
=
=
∵ 1000 mg = 1 g
So,
=
=
Thus,
"9.225 g" should be added.
Explanation:
Let us assume that total mass of the solution is 100 g. And, as it is given that acetic acid solution is 12% by mass which means that mass of acetic acid is 12 g and 88 g is the water.
Now, calculate the number of moles of acetic acid as its molar mass is 60 g/mol.
No. of moles =
=
= 0.2 mol
Molarity of acetic acid is calculated as follows.
Density =
1 g/ml =
volume = 100 ml
Hence, molarity =
=
= 2 mol/l
As reaction equation for the given reaction is as follows.
So, moles of NaOH = moles of acetic acid
Let us suppose that moles of NaOH are "x".
(as 1 L = 1000 ml)
x = 20 L
Thus, we can conclude that volume of NaOH required is 20 ml.