Chapter 6, Problem 6.61QP

Interpretation Introduction

Interpretation:

The molarity of the stock solution has to be determined.

Concept introduction:

The expression for molarity is given below.

    $M=\frac{molsolute}{Lsolution}$

It can be expressed in terms of mol solute by rearranging as below.

    $\text{mol}\text{solute}\text{=}\left(\text{M}\right)\left(\text{L}\text{solution}\right)$

The number of mol of solute should not change after dilution.

    $\begin{array}{c}{\text{mol}}_1\text{solute}\text{=}{\text{mol}}_2\text{solute}\\ \\ \left({\text{M}}_{\text{1}}\right)\left({\text{L}}_{\text{1}}\text{solution}\right)=\left({\text{M}}_2\right)\left({\text{L}}_2\text{solution}\right)\\ \\ \left({\text{M}}_{\text{1}}\right)\left(V_{\text{1}}\right)=\left({\text{M}}_2\right)\left(V_2\right)\end{array}$

That is the dilution formula is $\left({\text{M}}_{\text{1}}\right)\left(V_{\text{1}}\right)=\left({\text{M}}_2\right)\left(V_2\right).$  Where $\left({\text{M}}_{\text{1}}\right),\left({\text{M}}_2\right)$ are the molarity of the solution before and after dilution and $\left(V_{\text{1}}\right),\left(V_2\right)$ are the volume of the solution before and after dilution.