Assuming that the volume are additive, what is the concentration of KBr in a solution prepared by mixing 0.200 L of 0.053 M KBr with 0.550 L of 0.078 M KBr?

Given:

Volume of KBr solution(V₁) =$\text{0.200 L}$

Molarity of KBr solution(M₁)=$\text{0.053M}$

Volume of KBr solution(V₂) =$\text{0.550 L}$

Molarity of KBr solution(M₂)=$\text{0.078 M}$

$$\begin{gathered} \text{Molarity =}\frac{\text{moles of solute}}{\text{liters of solution}} \\ \text{moles of solute}=\text{Molarity}\times \text{liters of solution} \end{gathered}$$

$$\begin{gathered} \text{moles of KBr in 0.200 L of 0.53M} \\ \text{= 0.200 L}\times \text{0.53M} \\ \text{= 0.0106 moles} \\ \text{moles of KBr in 0.550 L of 0.078 M} \\ \text{= 0.550 L}\times \text{0.078M} \\ \text{= 0.0429 moles} \end{gathered}$$

Let us calculate the total number of moles present in solution after mixing.

$$\begin{gathered} \text{0.0106 moles}+\text{0.0429 moles} \\ 0.0535\text{moles} \end{gathered}$$

Let us calculate the total volume of solution after mixing.

$$\begin{gathered} \text{0.200 L}+\text{0.550 L} \\ \text{= 0.750 L} \end{gathered}$$

Let us calculate the concentration of solution after mixing.

$\text{Molarity =}\frac{\text{Total number of moles of solute}}{\text{Total volume of solution}}$

$$\begin{gathered} \text{Molarity =}\frac{0.0535\text{mole}}{\text{0.750 L}} \\ =0.0713\text{M} \end{gathered}$$

Result:

The concentration of solution obtained after mixing is $0.0713\text{M}$