Answered

At Westonci.ca, we connect you with the best answers from a community of experienced and knowledgeable individuals. Ask your questions and receive precise answers from experienced professionals across different disciplines. Get quick and reliable solutions to your questions from a community of experienced experts 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

Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.