Dilution of a Solution Strengths

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Discussion Overview

The discussion revolves around the mixing of solutions with different strengths, specifically how to create 6mg and 12mg strength solutions from a 0mg solution and an 18mg solution. The context includes mathematical reasoning related to solution dilution and volume calculations.

Discussion Character

  • Mathematical reasoning
  • Conceptual clarification
  • Homework-related

Main Points Raised

  • One participant seeks to understand how to mix 0mg and 18mg solutions to achieve 6mg and 12mg strengths, expressing confusion about the application of a dilution formula.
  • Another participant suggests a method for achieving the desired strengths, proposing that to create a 6mg solution, one should mix 2/3 of the 0mg solution with 1/3 of the 18mg solution, and for a 12mg solution, mix 1/3 of the 0mg with 2/3 of the 18mg.
  • A third participant confirms the reasoning of the second participant, providing a mathematical breakdown of how to achieve the 12mg solution by calculating the necessary volumes of each solution.
  • A later reply acknowledges the complexity of the calculations and expresses a sentiment of overcomplicating the math, indicating a personal struggle with the concepts involved.

Areas of Agreement / Disagreement

Participants generally agree on the proposed mixing ratios for achieving the desired solution strengths, though the initial confusion about the dilution formula indicates some uncertainty in the approach.

Contextual Notes

There is an assumption that the solution strengths are measured in milligrams per milliliter, and the discussion does not resolve potential ambiguities in the application of the dilution formula.

Who May Find This Useful

This discussion may be useful for individuals interested in solution preparation, particularly in fields such as chemistry or engineering where dilution and mixing of solutions are relevant.

frgeorgeh
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I want to mix 4 strengths of solutions. I thought to save time I would mix 1 a 0mg strength and a second at 18mg strength. How do I use those two solutions to make a 6mg, 12 mg strength solution?

The ratio of (OV)*(OS)=(NV)*(NS) where OV=Old Volume, OS=Old Strength,, NV=New Volume, NS=New Strength, does not work since I know all the results. For example, if I wanted to make the following:

I have: 1 gal (3,785ml) of 0mg and 1gal of 18mg

I want to make: 30ml bottles at 6mg and 12mg.

I hope this makes sense.

Thanks
 
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frgeorgeh said:
I want to mix 4 strengths of solutions. I thought to save time I would mix 1 a 0mg strength and a second at 18mg strength. How do I use those two solutions to make a 6mg, 12 mg strength solution?

The ratio of (OV)*(OS)=(NV)*(NS) where OV=Old Volume, OS=Old Strength,, NV=New Volume, NS=New Strength, does not work since I know all the results. For example, if I wanted to make the following:

I have: 1 gal (3,785ml) of 0mg and 1gal of 18mg

I want to make: 30ml bottles at 6mg and 12mg.

I hope this makes sense.

Thanks

__________________________________

OK, I think I am over thinking this. It seems to me, to get a 6mg strength, I add 2/3 of the 0MG and 1/3 of the 18mg. To get 12mg, I add 1/3 0mg and 2/3 of 18mg. What do u think?
 
I am assuming the solution strengths are in milligrams per milliliters. So, if you have a solution with a strength of $18$, to get a solution of volume $V$ with a strength of $18k$, where:

$$0\le k\le1$$

You would need to use $$kV$$ of the solution of strength $18$, with the rest consisting of the strength $0$ solution.

For example, since 12 is 2/3 of 18, then you would put 2/3 of 30 mL which is 20 mL with 10 mL of strength 0 to get 30 mL of strength 12.

So yes, your reasoning is correct. :D
 
MarkFL said:
I am assuming the solution strengths are in milligrams per milliliters. So, if you have a solution with a strength of $18$, to get a solution of volume $V$ with a strength of $18k$, where:

Thanks for the reply! I am an Engineer but always seem to over complicate my math.
 

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