Question

One antifreeze solution is 26% alcohol and another is 21% alcohol. How much of each mixture should be added to make 60 liters of a solution that is 24% alcohol?

Answer 1

Hint: When two or more substances combine to form a homogenous mixture, it is known as a solution. The formation of a solution depends on two factors, the solute, and the solvent, and whether they are miscible with each other.

Complete answer: Let us assume that the volume used of the solution containing 26% alcohol is ${{V}_{26}}$, the volume used of the solution containing 21% alcohol is ${{V}_{21}}$, and the volume of the resultant solution is ${{V}_{r}}$.
It is given to us that the solution needed to be prepared must have 24% alcohol and has a volume of 60 liters i.e., ${{V}_{r}}=60L.$
So, the concentration of alcohol in 60 liters 24% alcohol solution is
\[60\times \dfrac{24}{100}=14.4L\]
Now, according to the question, we need to take care of two conditions:
1. The alcohol content of the resultant solution must be equal to the alcohol content in the two solutions. This can be denoted by the equation
\[0.26{{V}_{26}}+0.21{{V}_{21}}=14.4\]
2. The volume after combining the two solutions must be equivalent to 60 liters.
\[{{V}_{26}}+{{V}_{21}}=60\]
So, from equation 1 and 2 we can say that
\[
   0.26{{V}_{26}}+0.21(60-{{V}_{26}})=14.4 \\
  \Rightarrow 0.26{{V}_{26}}+0.21\times 60-0.21{{V}_{26}}=14.4 \\
\Rightarrow 0.05{{V}_{26}}+12.6=14.4 \\
  \Rightarrow {{V}_{26}}=\dfrac{1.8}{0.05}=36L \\
\]
Since from equation 2, we know that
\[
  {{V}_{26}}+{{V}_{21}}=60 \\
  \Rightarrow {{V}_{21}}=60-{{V}_{26}} \\
\Rightarrow {{V}_{21}}=60-36 \\
\Rightarrow {{V}_{21}}=24L \\
\]
So, mixing 24 liters of 21% alcohol solution with 36 liters of 26% alcohol solution will give 60 liters of 24% solution.

Note: It should be noted that in this question, it was mentioned that the solution was an antifreeze solution just to confuse the student. There was no requirement or use of the nature of the solutions to solve the problem.