Chapter 20, Problem 91A

Interpretation Introduction

Interpretation: The amount of 6M $HCl$ solution needed to make $440mL$ of 1.5M $HCl$ solution is to be calculated.

Concept introduction: The molarity of a solution is defined as the number of moles of solute present per litre of a solution. So it details the concentration of solutions. That is the concentration of solutions will decrease with the addition of water.

The expression for molarity is as follows:

  $\text{M}=\frac{\text{moles of solute}}{\text{Volume of solution(L)}}$ …… (1)

Answer to Problem 91A

The volume of $HCl$ needed to make $440mL$ of 1.5M solutions from 6M $HCl$ solution is $\text{110}\text{mL}$ .

Explanation of Solution

The number of moles of solution needed to make $440mL$ (0.44L) of $HCl$ solution with a molarity of 1.5M can be calculated.

Substitute, $0.44L$ for volume of $HCl$ and $1.50\text{M}$ for molarity of solution in equation (1).

  $\begin{array}{c}\text{Moles solute}=\text{Volume }\left(\text{Litre}\right)\text{solution}\times \text{Molarity}\\ =1.5\times 0.44\\ =0.66\text{moles}\end{array}$

So $0.66\text{moles}$ of $HCl$ is needed to make 1.5M $HCl$ solution.

Now the volume of solution needed to make $0.66\text{moles}$ of $HCl$ from 6M solutions can be calculated.

Substitute, $0.44L$ for the volume of $HCl$ and $1.50\text{M}$ for molarity of solution in equation (1).

  $\begin{array}{c}\text{Volume}\left(\text{Litre}\right)\text{ solution}=\frac{\text{Moles solute}}{\text{Molarity}}\\ =\frac{0.66}{6}\\ =0.11\text{L}\\ \text{=110}\text{mL}\end{array}$

So, $\text{110}\text{mL}$ of 6M solutions is needed.