Chapter 6, Problem 6.102SP

Interpretation Introduction

Interpretation:

The volume (in millilitres) of $0.45{\text{ M BaCl}}_{\text{2}}$ that contains $15.0\text{ g}$ of ${\text{BaCl}}_{\text{2}}$ has to be calculated.

Concept Introduction:

Moles:

$\text{No}\text{. of moles (n) = }\frac{\text{mass}}{\text{Molar mass}}$

Molarity:

$\begin{array}{l}\text{Molarity = }\frac{\text{Moles of solute}}{\text{Litres of solution}}\\ \\ \text{Volume of solution = }\frac{\text{Moles of solute}}{\text{Molarity}}\end{array}$