Chapter 6.4, Problem 64STP

To determine

Tofind:thevolume of $\text{40\%}$ acid solution mixed with $\text{60\%}$ acid solutionto produce the $48\%$ concentration of acid solution.

Answer to Problem 64STP

  $80\text{ml}$ of $\text{40\%}$ acid solution should be mixed with $\text{60\%}$ acid solution to make the $48\%$ of acid solution having $200\text{ml}$ .

Explanation of Solution

Given:

One solution concentration is $60\%$ and another solution concentration is $40\%$ and the Concentration of total solution is $48\%$ and total volume is $200\text{ml}$ .

Concept used:

From a chemical reaction:

  $\text{Total volume of solution=}\frac{\text{volume of individual solution}}{\text{concentration}}.$

Calculation:

According to the given one solution concentration is $60\%$ and another solution concentration is $40\%$ and the concentration of total solution is $48\%$ and total volume is $200\text{ml}$ .

Total volume of the solution is equal to the ratio of volume of individual solution to the concentration of it.

Mathematically,

  $\text{total volume of solution=}\frac{\text{volume of individual solution}}{\text{concentration}}.$

  $\begin{array}{l}{v}_1=0.6\times 200,v_2=0.4\times 200.\\ v_1+v_2=\text{total volume}\left(v\right).\end{array}$

For $\text{40\%}$ acid solution the volume of the concentration will be:

  $\begin{array}{l}{v}_2=0.4\times 200.\\ v_2=80.\end{array}$

Hence. $80\text{ml}$ of $\text{40\%}$ acid solution should be mixed with $\text{60\%}$ acid solution to make the $48\%$ of acid solution having $200\text{ml}$ .