Chapter 4, Problem 4.44QE

(a)

Interpretation Introduction

Interpretation:

The molarity of $\text{250}\text{mL}$ solution prepared by dilution with water of $\text{0}\text{.12}\text{g}$ sodium nitrate has to be calculated.

Answer to Problem 4.44QE

The molarity of $\text{250}\text{mL}$ solution prepared by dilution with water of $\text{0}\text{.12}\text{g}$ sodium nitrate is $5.65\times 10^{-3}M.$

Explanation of Solution

Given: Volume of solution is $\text{250}\text{mL}$ ; Mass is $0.12\text{g}\text{.}$

Conversion of mass into moles:

  $\begin{array}{c}\text{Moles}\text{of}\text{solute}\text{=}\frac{\left(\text{Mass}\text{of}{\text{NaNO}}_3\right)}{\text{Molar}\text{mass}\text{of}{\text{NaNO}}_3}\\ \\ \text{=}\frac{0.12\text{g}}{85\text{g/mol}}\\ \\ \text{=}\text{0}\text{.0014}\text{mole}\text{.}\end{array}$

By substituting the given data into the molarity formula,

  $\begin{array}{c}\text{Molarity}\text{=}\frac{\text{Moles}\text{of}\text{solute}}{\text{Volume of solution in L}}\\ \\ =\frac{\text{0}\text{.0014}\text{mole}}{0.250L}=5.65\times 10^{-3}M.\end{array}$

Therefore,

The molarity of $\text{250}\text{mL}$ solution prepared by dilution with water of $\text{0}\text{.12}\text{g}$ sodium nitrate is $5.65\times 10^{-3}M.$

(b)

Interpretation Introduction

Interpretation:

The molarity of $\text{250}\text{mL}$ solution prepared by dilution with water of $\text{0}\text{.75}\text{mL}\text{of}\text{0}\text{.42}\text{M}\text{NaOH}$ has to be calculated.

Answer to Problem 4.44QE

The molarity of $\text{250}\text{mL}$ solution prepared by dilution with water of $\text{0}\text{.75}\text{mL}\text{of}\text{0}\text{.42}\text{M}\text{NaOH}$ is $1.26\times 10^{-3}M.$

Explanation of Solution

• Given information as follows,

Molarity and volume of $\text{NaOH}$ solution given as follows,

  $\begin{array}{l}{\text{M}}_{\text{1}}\text{=}0.42\text{M}\text{;}{\text{V}}_{\text{1}}\text{=}0.75m\text{L}\\ \\ {\text{M}}_{\text{2}}\text{=}\text{?}\text{;}{\text{V}}_{\text{2}}\text{=}\text{250mL}\end{array}$

Now apply ${\text{M}}_{\text{1}}{\text{V}}_{\text{1}}\text{=}{\text{M}}_{\text{2}}{\text{V}}_{\text{2}}$ relation.

By substituting into the relation, the volume is calculated as shown below,

  $\begin{array}{c}{\text{M}}_{\text{2}}\text{=}\frac{{\text{M}}_{\text{1}}{\text{V}}_{\text{1}}}{{\text{V}}_{\text{2}}}\\ \\ {\text{M}}_{\text{2}}\text{=}\frac{\left(\text{0}\text{.42}\text{M}\right)\left(\text{0}\text{.75}\text{mL}\right)}{\text{250}\text{mL}}\\ \\ \text{=}1.26\times 10^{-3}M.\end{array}$

Therefore,

The molarity of $\text{250}\text{mL}$ solution prepared by dilution with water of $\text{0}\text{.75}\text{mL}\text{of}\text{0}\text{.42}\text{M}\text{NaOH}$ is $1.26\times 10^{-3}M.$