Chapter 15, Problem 56A

Interpretation Introduction

Interpretation:The molarity of the final solution has to be calculated.

Concept Introduction: Molarity: Molarity is defined as the number of moles of solute in one liter of solution. Molarity is the preferred concentration unit for stoichiometry calculations. The formula is,

$\text{Molarity=}\frac{\text{Mass}\text{of}\text{solute(in}\text{moles)}}{\text{Volume}\text{of}\text{solution}\text{(in}\text{litres)}}$

Answer to Problem 56A

The molarity of the final solution is $\text{1}\text{.8}\text{M}\text{.}$

Explanation of Solution

Given,

Molarity of hydrochloric acid solution $\left(M_{\text{1}}\right)$ is $\text{2}\text{.5}\text{M}\text{.}$

Volume of hydrochloric acid solution $\left(V_{\text{1}}\right)$ is $\text{225}\text{.0}\text{mL}\text{.}$

Molarity of hydrochloric acid solution $\left(M_{\text{2}}\right)$ is $\text{0}\text{.75}\text{M}\text{.}$

Volume of hydrochloric acid solution $\left(V_{\text{2}}\right)$ is $\text{150}\text{.0}\text{mL}\text{.}$

Molarity is defined as the number of moles of solute in one liter of solution. The formula for calculating molarity is,

$\text{Molarity=}\frac{\text{Mass}\text{of}\text{solute(in}\text{moles)}}{\text{Volume}\text{of}\text{solution}\text{(in}\text{litres)}}$

Concentration of the solution is calculated using the formula,

$M_1{V}_1=M_2{V}_2$

Here,

$M_1\text{=}$ Initial molarity of the solution

$V_1=$ Initial volume of the solution

$M_2=$ Final molarity of the solution

$V_2=$ Final volume of the solution

The final molarity of the solution is calculated as,

Final molarity= $\frac{M_1{V}_1+M_2{V}_2}{V_1+V_2}$

Final molarity= $\frac{\left(\text{2}\text{.5}\text{M}\right)\left(\text{225}\text{.0}\text{mL}\right)\text{+}\left(\text{0}\text{.75}\text{M}\right)\left(\text{150}\text{.0}\text{mL}\right)}{\text{225}\text{.0}\text{mL+150}\text{.0}\text{mL}}$

Final molarity= $\frac{\text{562}\text{.5}\text{M+112}\text{.5}\text{M}}{\text{375}\text{mL}}$

Final molarity= $\text{1}\text{.8}\text{M}$

The molarity of the final solution is $\text{1}\text{.8}\text{M}\text{.}$

Conclusion

The molarity of the final solution is $\text{1}\text{.8}\text{M}\text{.}$