Chapter 19, Problem 129A

Interpretation Introduction

Interpretation : The number of liters of hydrochloric acid is to be calculated.

Concept Introduction : The number of moles of solute that are dissolved in one litre of solution is known as molarity. It also goes by the name "molar concentration." The formula used when the total number of moles are constant is given as

  ${\text{V}}_{\text{1}}\times {\text{M}}_{\text{1}}{\text{=V}}_{\text{2}}\times {\text{M}}_{\text{2}}$

Here, ${\text{V}}_{\text{1}}{\text{, M}}_{\text{1}}$ are the volume and molarity of initial solution and ${\text{V}}_{\text{2}}{\text{, M}}_{\text{2}}$ are the volume and molarity of final solution.

Answer to Problem 129A

There is about $0.47\text{L}$ needed to prepare $\text{HCl}$ solution.

Explanation of Solution

The concentration of an initial solution of $\text{HCl}$ is $\text{8}\text{.0}\text{M}$ .

The concentration of a final solution of $\text{HCl}$ is $2.5\text{M}$ .

The volume of the final solution of $\text{HCl}$ is $\text{1}\text{.50}\text{L}$ .

When the number of moles were constant, the formula is given as

  ${\text{V}}_{\text{1}}\times {\text{M}}_{\text{1}}{\text{=V}}_{\text{2}}\times {\text{M}}_{\text{2}}$

Here, ${\text{V}}_{\text{1}}{\text{, M}}_{\text{1}}$ are the volume and molarity of initial solution and ${\text{V}}_{\text{2}}{\text{, M}}_{\text{2}}$ are the volume and molarity of final solution.

To calculate the volume of initial solution, substitute the values in above formula:

  $\begin{array}{c}{\text{V}}_{\text{1}}\times {\text{M}}_{\text{1}}{\text{=V}}_{\text{2}}\times {\text{M}}_{\text{2}}\\ =\frac{{\text{V}}_{\text{2}}\times {\text{M}}_{\text{2}}}{{\text{M}}_1}\\ =\frac{1.50\text{L}\times 2.5\text{M}}{8.0\text{M}}\\ =0.47\text{L}\end{array}$

About $0.47\text{L}$ are needed to prepare the HCl solution.