Chapter 16, Problem 86E

Interpretation Introduction

Interpretation:

The volume of the $\text{9}\text{.76}\text{M}$ $\text{HCl}$ in $\text{mL}$ used for the preparation of $\text{2}\text{.00}\text{L}$ of $\text{0}\text{.400}\text{M}$ $\text{HCl}$ is to be calculated.

Concept introduction:

Molarity is defined as the number of moles of solute present in one liter of solution. This can be expressed by the relation given below.

$\text{M}=\frac{\text{n}}{\text{V}\left(\text{in L}\right)}$

On dilution, the molarity and the volume of the final solution are calculated by the relation given below.

${\text{M}}_{\text{c}}{\text{V}}_{\text{c}}={\text{M}}_{\text{d}}{\text{V}}_{\text{d}}$