Question 365004: pure acid is to be added to a 10% acid solution to obtain 90l of 75% solution? what amounts should each be used? Found 3 solutions by Fombitz, ewatrrr, josmiceli:Answer by Fombitz(32388) (Show Source):
You can put this solution on YOUR website! Let be the amount of 100% acid, the amount of 10% acid.
1.
.
.
2.
Subtract eq. 1 from eq. 2, liters
Then from eq. 1, liters
Hi,
Let x represent amount of pure acid(this will be an 100% solution) (90L-x the 10% solution
Write as we Read***
.10(90L-x) + 1.00x = .75*90L
solving for x
.90x = .65*90L
x = 65L, the amount of pure acid.
(90-65L)=25L the amount of the 10% solution
checking our answer
2.5 + 65 = 67.5
You can put this solution on YOUR website! Let = liters of pure acid to be added
Let = liters of 10% acid solution to be added
given:
(1)
(2)
----------------------
From (2):
(2)
(2)
Subtract (1) from (2)
(2)
(1)
And, since
65 liters of pure acid and 25 liters of 10% solution are needed
check answer:
OK
