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5 votes
Use a system of equations to solve this problem.

A chemist currently has two solutions of sodium chloride. One solution has a 5% concentration and the other has a 25% concentration. The chemist needs to make 10 L of a 10% sodium chloride solution.
Let x = the amount of 5% solution.
Let y = the amount of 25% solution.

How much of each solution does the chemist need to make?
Enter your answers, as decimals, in the boxes.
____ L of 5% solution

____ L of 25 % solution

2 Answers

2 votes
We have 5% and 25% solutions
We need 10 liters of 10% solution.
A) x + y = 10
B) .05x + .25y = (10*.10)
Multiply A) by -.25
A) -.25x -.25y = -2.50 then adding to B)
B) .05x + .25y = (10*.10)
-.2x = -1.50
x = 7.5
y = 2.5

User BitTickler
by
6.3k points
5 votes

Answer:

The 7.5 L of 5% solution and 2.5 L of 25 % solution .

Explanation:

As given

A chemist currently has two solutions of sodium chloride.

One solution has a 5% concentration and the other has a 25% concentration.

The chemist needs to make 10 L of a 10% sodium chloride solution.

Let x = the amount of 5% solution.

Let y = the amount of 25% solution.

Than first equation becomes

x + y = 10

5% is written in the decimal form .


= (5)/(100)

= 0.05

25% is written in the decimal form .


= (25)/(100)

= 0.25

10% is written in the decimal form .


= (10)/(100)

= 0.10

Than the second equations becomes

Concentration of 5% solution × Amount of solution + Concentration of 25% solution × Amount of solution = Concentration of 10% solution × Amount of solution .

Putting all the values in the above

0.05x + 0.25y = 10 × 0.10

Simplify the equation


(5x)/(100)+(25y)/(100) = (10* 10)/(100)

5x + 25y = 100

Than two equations are

x + y = 10

5x + 25y = 100

Multiply x + y = 10 by 5 and subtracted from 5x + 25y = 100 .

5x - 5x + 25y - 5y = 100 - 50

20y = 50


y = (50)/(20)

y = 2.5 L

Putting the value of y in the equation

x + y = 10

x + 2.5 = 10

x = 10 - 2.5

x = 7.5 L

Therefore the 7.5 L of 5% solution and 2.5 L of 25 % solution .



User Aaronmallen
by
7.1k points
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