Question

A biologist has two brine solutions, one containing 4% salt and another containing 16% salt. How many milliliters of each solution should she mix to obtain 1 L of a solution that contains 11.2% salt?

Answer 1

Answer:

400 mL of 4% salt solution and 600 mL of 16% salt solution need to be mixed

Explanation:

Let's assume concentration of brine salt solutions are in %(w/v) unit.

Final mixture has a concentration of 11.2% salt and volume of 1 L or 1000 mL

Hence, amount of salt in final mixture = $\frac{11.2\times 1000}{100}g=112g$

Suppose x mL of 4% salt solution and (1000-x) mL of 16% salt solution need to be mixed to get final mixture

So, amount of salt in x mL of 4% salt solution = $\frac{4x}{100}g$

      amount of salt in x mL of 16% salt solution =$\frac{16\times (1000-x)}{100}g$

Hence, $\frac{4x}{100}+\frac{16\times (1000-x)}{100}=112$

or, $x=400$

So, 400 mL of 4% salt solution and 600 mL of 16% salt solution need to be mixed