Chapter 14, Problem 91AE

Interpretation Introduction

Interpretation:

Grams of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ formed by reaction of $80.5\text{ mL}$ of $0.642M\text{ Ba}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ with $44.4\text{ mL}$ of $0.743M\text{ KOH}$ have to be determined.

Concept Introduction:

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

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