How many mL of 50% (v/v) alcohol should be mixed with 10% alcohol to obtain 50mL of 35% (v/v) alcohol? Round to the hundredths place.

The volume of 50% alcohol used to make the mixture = 31.25 L

The volume of 10% alcohol used to make the mixture = 18.75 L

Explanation:

Let the volume of 50% alcohol used to make the mixture = x L

Let the volume of 10% alcohol used to make the mixture = y L

Total volume of the mixture = x + y = 50 L ... (1)

For 50% alcohol:

C₁ = 50%, V₁ = x L

For 10% alcohol:

C₂ = 10%, V₂ = y L

For the resultant alcohol solution:

C₃ = 35%, V₃ = 50 L

Using

C₁V₁ + C₂V₂ = C₃V₃

50*x + 10*y = 35*50

So,

5x + y = 175 ... (2)

Solving 1 and 2 we get,

x = 31.25 L

y = 18.75 L