Question

How many liters of a $ 90\% $ solution of concentrated acid needs to be mixed with a $ 75\% $ solution of concentrated acid to get a $ 30L $ solution of $ 78\% $ concentrated acid?
(A) $ 24L $
(B) $ 22.5L $
(C) $ 6L $
(D) $ 17.5L $

Answer 1

Hint: By substituting the values in the weighted average formula, the volume of a final solution that was mixed from two different percentages can be determined by considering the total percent as $ 100\% $ consider the given percentages as normality and the volume be noted as $ V $
 $ {N_1}V + {N_2}\left( {30 - V} \right) = N \times 30 $
 $ {N_1} $ is the first given percent
 $ {N_2} $ is the second given percent
 $ N $ is the final percent obtained solution
 $ V $ be the volume.

Complete Step By Step Answer:
Given that some volume of a $ 90\% $ solution of concentrated acid needs to be mixed with a $ 75\% $ solution of concentrated acid to get a $ 30L $ solution of $ 78\% $ concentrated acid.
As the solution is given in terms of percent, substitute the percentages in the place of normality, as both are the terms used to express the concentration only.
The volume of solution that has to be prepared is $ 30L $
Let us consider the $ V $ litre of $ 90\% $ solution of concentrated acid was mixed and $ \left( {30 - V} \right) $ litre of $ 75\% $ solution of concentrated acid was mixed.
According to the formula,
 $ 90 \times V + 75\left( {30 - V} \right) = 78 \times 30 $
By simplifying the above equation,
 $ V = 6litres $
Thus, $ 6L $ of a $ 90\% $ solution of concentrated acid needs to be mixed with a $ 75\% $ solution of concentrated acid to get a $ 30L $ solution of $ 78\% $ concentrated acid.

Note:
As the number of moles in both the solutions before and after the mixing of solutions are the same. By the definition of normality, the number of moles is equal to the product of normality and volume. Thus, the above formula was used.