Answer:
Part 1: 89 mL of 26% solution
Part 2: 80 mL of 13% solution
Step-by-step explanation:
Let's break this problem down step by step!
1. Let x be the amount of 26% solution needed.
2. Since the total volume of the mixture is 169 mL, the amount of 13% solution needed is 169 - x.
3. The total amount of saline in the mixture is 20% of 169 mL, which is 0.20 × 169 = 33.8 g.
4. The amount of saline in the 26% solution is 0.26x, and the amount of saline in the 13% solution is 0.13(169 - x).
5. Set up the equation: 0.26x + 0.13(169 - x) = 33.8.
Solving for x, we get:
0.26x + 22.17 - 0.13x = 33.8
0.13x = 11.63
x ≈ 89.46 mL
So, approximately 89 mL of the 26% solution is needed.
To find the amount of 13% solution needed, subtract x from 169:
169 - 89 ≈ 80 mL
Therefore, the pharmacist should mix approximately 89 mL of the 26% solution with 80 mL of the 13% solution.
Rounding to the nearest mL, the answers are:
Part 1: 89 mL of 26% solution
Part 2: 80 mL of 13% solution