Chapter 15.2, Problem 7RQ

Interpretation Introduction

Interpretation: The new concentration when $\text{50}\text{mL}$ of water is added to $\text{725}\text{mL}$ of $\text{1}\text{.25}\text{M}\text{NaCl}$ has to be determined.

Concept introduction: Dilute solutions are prepared by adding more water or solvent to a specified amount of solutions. During dilution of any specified amount of solvent, only solvent or water is added into the solution. The quantity of solute remains intact before and after the dilution in solution. This is the key to obtain desired concentration of solution through dilution. Formula for dilution is as follows:

  ${\text{M}}_1{\text{V}}_{\text{1}}={\text{M}}_2{\text{V}}_2$

Where,

  ${\text{M}}_1$ and ${\text{M}}_2$ is initial and final molarity of solution.

  ${\text{V}}_{\text{1}}$ and ${\text{V}}_2$ is initial and final volume of solution.

Answer to Problem 7RQ

The new concentration when $\text{50}\text{mL}$ of water is added to $\text{725}\text{mL}$ of $\text{1}\text{.25}\text{M}\text{NaCl}$ is $0.0862\text{M}$ .

Explanation of Solution

Given information: Initial volume and final volume is $\text{50}\text{mL}$ and $\text{725}\text{mL}$ . Intial molarity of solution is $\text{1}\text{.25}\text{M}$ .

Formula for dilution is as follows:

  ${\text{M}}_1{\text{V}}_{\text{1}}={\text{M}}_2{\text{V}}_2$

Rearrange above equation to obtain new molarity of solution.

  ${\text{M}}_2=\frac{{\text{M}}_1{\text{V}}_{\text{1}}}{{\text{V}}_2}$

Substitute $\text{50}\text{mL}$ for ${\text{V}}_{\text{1}}$ , $\text{725}\text{mL}$ for ${\text{V}}_2$ and $\text{1}\text{.25}\text{M}$ for ${\text{M}}_1$ in above equation.

  $\begin{array}{c}{\text{M}}_2=\frac{{\text{M}}_1{\text{V}}_{\text{1}}}{{\text{V}}_2}\\ =\frac{\left(\text{1}\text{.25}\text{M}\right)\left(\text{50}\text{mL}\right)}{\left(\text{725}\text{mL}\right)}\\ =0.0862\text{M}\end{array}$

Conclusion

The new concentration when $\text{50}\text{mL}$ of water is added to $\text{725}\text{mL}$ of $\text{1}\text{.25}\text{M}\text{NaCl}$ is $0.0862\text{M}$ .