The volume of methanol necessary to prepare the antifreeze good for antifreeze solution is 3.2 L
M₁V₁ = M₂V₂
Where
The volume of the methanol necessary to prepare the solution can be obtained as illustrated below:
M₁V₁ = M₂V₂
24.7 × V₁ = 10 × 8
24.7 × V₁= 80
Divide both side by 24.7
V₁ = 80 / 24.7
V₁ = 3.2 L
Thus, the volume of methanol necessary to prepare the antifreeze good for antifreeze solution is 3.2 L
Learn more about dilution:
https://brainly.com/question/15022582
#SPJ1