Chapter 14, Problem 33PE

(a)

Interpretation Introduction

Interpretation:

Molarity of solution prepared by addition of $125\text{ mL}$ of $6.0M{\text{ HC}}_2{\text{H}}_3{\text{O}}_2$ with $525\text{ mL}$ of ${\text{H}}_2\text{O}$ has to be determined.

Concept Introduction:

Molarity is amount of solute present in $1\text{ L}$ of solution. Formula to calculate molarity of solution is as follows:

  $\text{Molarity of solution}=\frac{\text{Moles of solute}}{\text{Volume }\left(\text{L}\right)\text{ of solution}}$

(b)

Interpretation Introduction

Interpretation:

Molarity of solution prepared by addition of $125\text{ mL}$ of $6.0M{\text{ HC}}_2{\text{H}}_3{\text{O}}_2$ with $\text{175 mL}$ of $\text{1}\text{.5 }M{\text{ HC}}_2{\text{H}}_3{\text{O}}_2$ has to be determined.

Concept Introduction:

Molarity is amount of solute present in $1\text{ L}$ of solution. Formula to calculate molarity of solution is as follows:

  $\text{Molarity of solution}=\frac{\text{Moles of solute}}{\text{Volume }\left(\text{L}\right)\text{ of solution}}$