Chapter 4, Problem 4.11P

(a)

Interpretation Introduction

Interpretation:

The meaning of each symbol and the reason for the following equation to work is to be determined.

$M_{\text{dil}}{V}_{\text{dil}}=M_{\text{conc}}{V}_{\text{conc}}$

Concept introduction:

Dilution is the process of converting a concentrated solution into a dilute solution by adding the solvent. In the resultant solution, the volume of the solution increases due to the addition of the solvent.

The expression to calculate the number of moles is as follows:

$\text{moles of solute}=\left(\text{molarity of solution}\right)\left(\text{volume of solution}\right)$

(b)

Interpretation Introduction

Interpretation:

The expression to calculate the amount (mol) and the mass (g) of solute when the volume and molarity of a ${\text{CaCl}}_2$ solution are given is to be determined.

Concept introduction:

Molarity $\left(M\right)$ is one of the concentration terms that determine the number of moles of solute present in per liter of solution. Unit of molarity is $\text{mol/L}$.

The expression to calculate the moles of the compound when molarity of solution and volume of solution are given is as follows:

$\text{Moles of compound}\left(\text{mol}\right)=\left[\begin{array}{c}\text{volume of solution}\left(\text{L}\right)\\ \left(\frac{\text{molarity}\text{of}\text{solution}\left(\text{mol}\right)}{\text{1L of solution}}\right)\end{array}\right]$

The expression to calculate the amount of compound when moles of the compound and molecular mass are given:

$\text{amount of compound}\left(\text{g}\right)=\text{moles of compound}\left(\text{mol}\right)\left(\frac{\text{molecular mass of compound}\left(\text{g}\right)}{1\text{mole of compound}}\right)$