Asked by amy
A lab assistant wants to make FIVE LITERS of 31.2% acid solution. If solutions of 40% and 18% are in stock, how many liters of each must be mixed to prepare the solution.
i don't even know what equation to use or anything, please help me, thank you.
i don't even know what equation to use or anything, please help me, thank you.
Answers
Answered by
Steve
the amount of acid in a solution is the % times the volume.
For example 5L of 50% solution has 2.5L of acid.
So, if you add up all the acid in the parts of your solution, they must add up to the total acid at the end.
If x is the amount of 18% solution, then there must be 5-x liters of 40% solution, since the total volume is 5 liters.
Now, adding up the amount of acid, we need
.18x + .40(5-x) = .312(5)
.18x + 2 - .4x = 1.56
.22x = .44
x = 2
so, 2L of 18% and 3L of 40% will produce 5L of 31.2%
.18*2 + .40*3 = .36+1.2 = 1.56 = .312*5
For example 5L of 50% solution has 2.5L of acid.
So, if you add up all the acid in the parts of your solution, they must add up to the total acid at the end.
If x is the amount of 18% solution, then there must be 5-x liters of 40% solution, since the total volume is 5 liters.
Now, adding up the amount of acid, we need
.18x + .40(5-x) = .312(5)
.18x + 2 - .4x = 1.56
.22x = .44
x = 2
so, 2L of 18% and 3L of 40% will produce 5L of 31.2%
.18*2 + .40*3 = .36+1.2 = 1.56 = .312*5
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