Help! I cant solve this

Question

You need a 30% alcohol solution. On hand, you have a 270 mL of a 45% alcohol mixture. How much pure water will you need to add to obtain the desired solution?

Answer 1

Hello, thank you for taking the time to post your question!

The key on solving this type of word question is to use the information given in order to set up an equation that represents the situation. For this example that would be

Alcohol = V(initial) x C(initial) = 270 mL x 45% = 121.5 mL Alcohol

So that will be the amount of Alcohol present in the end

Then we want to take

V(final) = Amount of Alcohol / C(final) = 121.5 / (0.30) = 405 mL

So if the end solution is 405 mL total and 270 mL is the alcohol mixture we just want to subtract to find the amount of water

V(water) = 405 – 270 = 135 mL

What that means is that in order to obtain the desired solution you need to add 135 mL of pure water.

I hope that helps get you moving in the right direction! Feel free to reach out if you have any additional questions beyond that :)

Answer 2

Please see my solution to your other similar problem.

The generic process:

1. find the amount of each constituent in the solution.
2. Assuming that the metric is the % of alcohol, then one of the following applies:
3. to increase concentration, add alcohol
4. to DECREASE concentration, add water
5. Since you are decreasing the concentration by adding water, the total amount of pure alcohol remains constant. Using this, and the desired new concentration, find the total amount of water in the new mix.
6. Finally, find the difference between the existing water volume and the desired total. That is what needs to be added.

You need a 30% alcohol solution. On hand, you have a 270 mL of a 45% alcohol mixture. How much pure water will you need to add to obtain the desired solution?

The starting solution has 121.5ml pure alcohol (0.45 * 270), and 270 - 121.5 = 148.5ml water

For a 30% solution, the water volume will be 7/3 times the alcohol, so new water amount will be 121.5 * 7/3 = 283.5ml.

The amount of water added must then be 135ml

Answer 3

Suppose you add W mL of pure water. Combine with the alcohol mixture and you will have 270 + W mL

Now consider how much alcohol is added together to make the desired solution. The 270 mL is 45% alcohol; the W mL is 0% alcohol; and the final solution of 270 + W should be 30% alcohol.

0.45*(270) + 0*(W) = 0.30*(270 + W)

121.5 + 0 = 81 + 0.30W

40.5 = 0.30W

135 = W

So add 135 mL of pure water.

Answer 4

Let W = ml of water to add

..45*270/(270+W)=.3

.15*270=W=40.5 ml

Explanation:

270 ml of 45% alcohol contains .45*270 "pure" alcohol.

This amount of "pure" alcohol is diluted into 270+W ml of solution

The concentration of the final solution is to be 30%, i.e. .45*270/(270+W)=.3