Answered

Welcome to Westonci.ca, the place where your questions are answered by a community of knowledgeable contributors. Our platform connects you with professionals ready to provide precise answers to all your questions in various areas of expertise. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.

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

We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.