SOLUTION: A chemist has two solutions of HNO3. One has a 40% concentraiton and the other has a 25% concentration to obtain 58 liters of a 35% solution. How many liters of each solution must
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Question 1032589: A chemist has two solutions of HNO3. One has a 40% concentraiton and the other has a 25% concentration to obtain 58 liters of a 35% solution. How many liters of each solution must be mixed to obtain 58 liters of a 35% solution? Answer by Edwin McCravy(20056) (Show Source):
Let the number of liters of the stronger solution be x
Let the number of liters of the weaker solution be y
% pure Liters
Type Liters of of
of of each PURE
solution liquid solution HNO3
-------------------------------------------
stronger x 0.40 0.40x
weaker y 0.25 0.25y
-------------------------------------------
mixture 58 0.35 0.35(58) = 20.3 liters
The first equation comes from the liters of liquid column.
x + y = 58
The second equation comes from the last column.
0.40x + 0.25y = 20.3
Get rid of decimals by multiplying every term by 100:
40x + 25y = 2030
So we have the system of equations:
.
We solve by substitution. Solve the first equation for y:
x + y = 58
y = 58 - x
Substitute (58 - x) for y in 40x + 25y = 2030
40x + 25(58 - x) = 2030
40x + 1450 - 25x = 2030
15x + 1450 = 2030
15x = 580
x = 38.6666667 = the number of liters of the stronger.
Substitute in y = 58 - x
y = 58 - (38.6666667)
y = 19.3333333 = the number of liters of the weaker.
Edwin