Question

A chemist needs 90 milliliters of a 72% solution but has only 68% and 77% solutions available. Find how many milliliters of the 68% solution should be mixed with the 77% solution to get the desired strength.

Answer 1

Mix 50 mL of the 68% solution with 40 mL of the 77% solution to make 90 mL of the 72% solution.

Step-by-step explanation:

Let the volume required of the 68% solution be
`x` mL. The volume of the 68% solution and the 77% solution shall add up to 90 mL. Thus the volume required of the 77% solution shall be
`(90- x)` mL.

The amount of solute in the 72% solution will be:

`90* 72\% = 64.8`.

The
`x` mL of the 68% solution will contribute:

`68\% \cdot x = 0.68\;x`.

The
`(90- x)` mL of the 77% solution will contribute:

`77\% \cdot x = 0.77\;(90 - x) = 69.3 - 0.77\;x`.

The two values shall add up to
`64.8`. That is:

`0.68\;x + 69.3 - 0.77\;x = 64.8`.

`-0.09\;x = -4.5`.

`\displaystyle x = (4.5)/(0.09) = 50`.

In other words, there need to be

• 
  `\rm 50\;mL` of the 68% solution, and
• 
  `\rm 90 - 50 = 40\;mL` of the 77% solution.