Question

How many liters of each solution are mixed when \[90\% \] acid solution ( \[90\% \] pure acid and \[10\% \] water) is mixed with \[97\% \] acid solution to obtain \[21litres\] of \[95\% \] acid solution.

Answer 1

Hint: Concentration of a solution is the measure of the amount of solute contained in a fixed amount of solvent. Concentrations can be expressed by various methods and one such method is volume by volume percentage which involves the volume of solute and the solvent.

Complete answer: The given percentages are the concentrations in volume by volume percentages. Hence a \[95\% \] solution means that ninety five (units of volume) of acids are dissolved in five (units of volume) of water making up a total volume of hundred (units of volume).
Since the question deals with litres of volume, we will keep the same units to make our calculation easier.
Now, let us assume that the total volume of the first solution is \[{V_1}\] i.e. we have \[{V_1}\] litres of \[90\% \] acid solution.
Also, let us assume that the total volume of the second solution is \[{V_2}\] i.e. we have \[{V_2}\] litres of \[97\% \] acid solution.
The volume of acid in any particular solution can be calculated using the following formula:
\[{\text{volume of acid}} = {\text{volume percentage}} \times {\text{total volume of the solution}}\]
Thus the volume of acid present in \[90\% \] solution will be: \[\dfrac{{90}}{{100}} \times {V_1}\]
And the volume of acid present in \[97\% \] solution will be: \[\dfrac{{97}}{{100}} \times {V_2}\]
And the volume of acid present in \[21litres\] of \[95\% \] solution will be: \[\dfrac{{95}}{{100}} \times 21 = \dfrac{{1995}}{{100}}litres\]
The volume of water in each solution will be the remaining part of the percentage.
Thus the volume of water present in \[90\% \] solution will be: \[\dfrac{{10}}{{100}} \times {V_1}\]
And the volume of water present in \[97\% \] solution will be: \[\dfrac{3}{{100}} \times {V_2}\]
And the volume of acid present in \[21litres\] of \[95\% \] solution will be: \[\dfrac{5}{{100}} \times 21 = \dfrac{{105}}{{100}}litres\]
The net volume of acid and water upon mixing will be the same as that contained in the mixed solution. This observation gives rise to two important equations:
\[\left( {\dfrac{{10}}{{100}} \times {V_1}} \right) + \left( {\dfrac{3}{{100}} \times {V_2}} \right) = \dfrac{{105}}{{100}}litres{\text{ (1)}}\]
\[\left( {\dfrac{{90}}{{100}} \times {V_1}} \right) + \left( {\dfrac{{97}}{{100}} \times {V_2}} \right) = \dfrac{{1995}}{{100}}litres{\text{ (2)}}\]
The above two equations are linear equation in two variable, upon solving equations \[{\text{(1)}}\] and \[{\text{(2)}}\] we get,
\[{V_1} = 6litres\]
\[{V_2} = 15litres\]
Hence the volumes of first and second solution that are mixed are \[{V_1} = 6litres\] and \[{V_2} = 15litres\].

Note:
The answer can be confirmed by putting these volumes into their respective equations, if the equations get balanced (i.e. the left hand side becomes equal to the right hand side) then the values are correct. Also, the sum of these volumes comes out to be equal to the total volume of mixture which confirms that the answers are correct.