Mixture problems- Pure Solutions

Question

You need a 25% alcohol solution. On hand, you have a 300 mL of a 40% alcohol mixture. How much pure water will you need to add to obtain the desired solution?

Ive tried several times and can't seem to get the hang of this one, thanks!

Answer 1

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With problems like this, ask yourself what component doesn’t change. In this case, since you are only adding water, the absolute amount of alcohol doesn’t change.

You have 300 mL of a 40% solution, so you have 0.40 x 300 mL = 120mL of alcohol which doesn’t change when diluting with water.

The final volume of solution you will have after adding water can be written V + 300 mL where V is the amount of water to be added.

The final solution will be 25 % alcohol, so you can write 0.25 x (V + 300) = 120 mL because the volume of alcohol doesn’t change.  You now have only one unknown, V, so you can solve the equation for V.

Hope this helps.

Answer 2

If the original solution is 40% alcohol, then it is 60% water. Water is, of course, 100% water, and the mixture is 25% alcohol (or 75% water).

Let x = number of ml of pure water to be added

Then, 0.60(300) + 1.00x = 0.75(x + 300)

180 + x = 0.75x + 225

0.25x = 45

x = 180 ml

Answer 3