Question

You need 180 mL of a 25% alcohol solution. On hand, you have a 30% alcohol mixture. How much of the 30% alcohol mixture and pure water will you need to obtain the desired solution?

Answer 1

Answer:

150 mL of the 30% alcohol mixture and 30 mL of the pure water will we need to obtain the desired solution.

Explanation:

Let the volume of 30% alcohol used to make the mixture = x L

For 25% alcohol:

C₁ = 25% , V₁ = 180 mL

For 30% alcohol :

C₂ = 30% , V₂ = x L

Using  

C₁V₁ = C₂V₂

25×180 = 30×x

So,  

x = 150 mL

Pure water = 180 mL - 150 mL = 30 mL

150 mL of the 30% alcohol mixture and 30 mL of the pure water will we need to obtain the desired solution.