# Dilution of a Solution Strengths

#### frgeorgeh

##### New member
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

#### frgeorgeh

##### New member
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?

#### MarkFL

Staff member
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:

$$\displaystyle 0\le k\le1$$

You would need to use $$\displaystyle kV$$ 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.

#### frgeorgeh

##### New member
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.
The text passage contains a mathematical question about diluting solutions to achieve specific strengths, and an answer is provided in the text.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Dilution of a Solution Strengths

#### frgeorgeh

##### New member
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

#### frgeorgeh

##### New member
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?

#### MarkFL

Staff member
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:

$$\displaystyle 0\le k\le1$$

You would need to use $$\displaystyle kV$$ 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.

#### frgeorgeh

##### New member
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.
user
The text passage contains a mathematical question about diluting solutions to achieve specific strengths, and an answer is provided in the text.

TYPE: SINGLE_QUESTION_WITH_ANSWER