Question

A biologist has two brine solutions, one

containing 5% salt and another containing 20% salt. How

many milliliters of each solution should she mix to obtain

1 L of a solution that contains 14% salt?

Short answer

For the given mixture problem, we get that 400 ml of 5% salt solution should be mixed with 600 ml of 20% salt solution.

Step-by-step solution

Step 1. Given information.

x = concentration of sulphuric acid in the first container.

y = concentration of sulphuric acid in the second container.

Step 2. Write down the concept.

Formulating a table for the given information dividing the bkending procedures as case 1 and case 2, we get,

300x + 600y =135 …….(1)

100x + 500y = 75 ……..(2)

Step 3. Determining the values

Solving equation by elimination method we have,

3 $\times$ equation(2) gives 300x + 1500y – 225

Now solving,

300x + 600y = 135

-(300x +1500y = 225)

We get,

900y = 90

y = 0.1

y =10%

Substituting in (1) we get

300x + 600 $\times$ 0.1 = 135

300x = 75

x = 0.25

x = 25%