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OleMash [197]
3 years ago
15

A solution that is 20% acid and 80% water is mixed with a solution that is 50% acid and 50% water. If twice as much 50% acid sol

ution is used as 20% solution, then what is the ratio of acid to water in the mixture of the solutions?
Chemistry
1 answer:
eduard3 years ago
7 0

Answer:

The ratio acid to water in the mixture is 2:3

Explanation:

Let the volume of 20% acid solution used to make the mixture = x units

So, the volume of 50% acid solution used to make the mixture = 2x units

Total volume of the mixture = x + 2x = 3x units

For 20% acid solution:

C₁ = 20% , V₁ = x

For 50% acid solution :

C₂ = 50% , V₂ = 2x

For the resultant solution of sulfuric acid:

C₃ = ? , V₃ = 3x

Using

<u>C₁V₁ + C₂V₂ = C₃V₃</u>

20×x + 50×2x = C₃×3x

So,

20 + 50×2 = C₃×3

Solving

120 = C₃×3

C₃ = 40 %

Thus, for the resultant mixture,

<u>Acid percentage = 40%</u>

<u>Water percentage = (100 - 40)% = 60%</u>

<u>Ratio acid to water in the mixture = 40:60 = 2:3</u>

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