Chemistry
Mixture
Problems
A chemist needs 150 milliliters of a 50%
saline solution but has only 18% and 78%
solutions available. Find how many
milliliters of each that should be mixed to
get the desired solution.
We’re asked to find the amount used of
each of the solutions she has on hand, so …
Let x = the mL of the 18% solution
Let y = the mL of the 78% solution
Now we need to find out two relationships between these amounts:
1) A chemist needs 150 milliliters of solution.
The amount of 18% solution + the amount of 78% solution = 150 mL
Our first equation is x + y = 150
We need one more relationship between the amounts:
2) A chemist needs 150 milliliters of a 50% saline solution.
That means that she needs 150(.05) or 75 mL of salt in her solution
Amount of salt in the 18% solution = .18x
Amount of salt in the 78% solution = .78y
Our second equation is .18x + .78y = 75
Now we have two equations in two unknowns
x + y = 150
.18x + .78y = 75
I would solve this by elimination
x + y = 150
.18x + .78y = 75 multiply by 100 to clear the decimals
x + y = 150
18x + 78y = 7500
Multiply the top equation by -18 to clear the x column:
(I picked 18 because it’s smaller than 78 - the numbers will be easier
to work with)
x + y = 150
18x + 78y = 7500
-18x - 18y = -2700
18x + 78y = 7500
Now just add the equations together. The x’s will add up to zero and
disappear!
60y = 4800
y = 80
She needs 80ml of 78% solution.
Now let’s find the amount of 18% solution and check our work
x + y = 150
y = 80
x + 80 = 150
x = 70
Now we have 80 mL of 78% solution and 70 mL of 18% solution.
Does this fit our story?
Does 80 at 78% + 70 at 18% = 75?
80(.78) + 70(.18)
62.4+ 12.6 = 75
We have a winner!
80 ml of 78% solution and 70 mL of 18% solution