8L of a 45% saline solution was mixed with 4L of pure water. What is the concentration of the mixture?

Question

I forgot how to find concentration

Answer 1

Saline Solution Mixture Problem

Question: 8L of a 45% saline solution was mixed with 4L of pure water. What is the concentration of the mixture?

Solution

This is a classic percent mixture problem. The general formula for this type of problem is Q = Ar, where:

1. Q is the quantity of a substance in some solution
2. r is the percent of concentration
3. A is the amount of the solution

Problem Solving Strategy:

1. Define the unknown quantityLet r = the percent concentration of the mixture
2. (In decimal form. Remember: 100%r = final answer.)
3. Apply Q = Ar to each solution We can record the results in tabular format:

SolutionAmount of solution (A) in LPercent of Concentration (r)Quantity of substance (Q) in L
Saline Water	8	0.45	0.45(8)
Pure Water	4	0	0(4)
Mixture	(8+4) = 12	r	(8+4)r

1. Determine the relationship between quantitiesThe sum of the quantities of the substances being mixed must equal the quantity of the substance after mixing.
2. Set up and solve the equation0.45(8) + 0 = (8+4)r
3. 0.45(8) = 12r
4. (Multiply both sides by 100 to clear the decimal)
5. 45(8) = 1,200r
6. 360 = 1,200r
7. 360/1,200 = r
8. (Reduce the fraction to simplest terms by dividing numerator and denominator by the GCF of 120)
9. 360/1,200 = 3/10 = 30/100 = 30%

Therefore, the mixture is 30% saline.

Verification

Let's verify our answer:

1. [0.45(8) + 0(4)] L = [0.30(8+4)] L ?
2. (3.6 + 0) L = 0.30(12) L ?
3. 3.6 L = 3.6 L ✓