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A pharmacist wants to mix a 26% saline solution with a 13% saline solution to get 169 mL of a 20% saline solution. How much of each solution should be
used? Round your answer to the nearest mL.
Part 1 of 2
mL of 26% solution should be mixed.
Part 2 of 2
mL of 13% solution should be mixed.

Sagot :

subhamjrp

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

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