Chapter 5, Problem 61A

(a)

Interpretation Introduction

Interpretation: The volume of 50.0 g sodium chloride to be calculated.

Concept introduction: Amount of matter present in a substance in given volume is density. That is, density is ratio of mass of an object to its volume or mass per unit volume.

Answer to Problem 61A

Volume of sodium chloride is 23.1mL

Explanation of Solution

Expression for density is as follows:

  $\text{Density=}\frac{\text{Mass}}{\text{Volume}}$

Given, density of sodium chloride is $\text{2}\text{.16}\text{g}/{\text{cm}}^{\text{3}}$ and mass is 50.0g. Put these values in density expression is as follows:

  $\begin{array}{c}\text{2}\text{.16}\text{g}/{\text{cm}}^{\text{3}}\text{=}\frac{\text{50}\text{.0g}}{\text{Volume}}\\ \text{Volume=}\frac{\text{50}\text{.0g}}{2.16\text{g}/{\text{cm}}^{\text{3}}}\\ \text{=23}{\text{.1cm}}^{\text{3}}\\ =23.1\text{mL}\end{array}$

Thus volume of sodium chloride is 23.1mL

(b)

Interpretation Introduction

Interpretation: The volume of 50.0 g mercury to be calculated.

Concept introduction: Amount of matter present in a substance in given volume is density. That is, density is ratio of mass of an object to its volume or mass per unit volume.

Answer to Problem 61A

Volume of mercury is 3.68mL.

Explanation of Solution

Expression for density is as follows:

  $\text{Density=}\frac{\text{Mass}}{\text{Volume}}$

Given, density of mercury is $\text{13}\text{.6}\text{g}/{\text{cm}}^{\text{3}}$ and mass is 50.0g. Put these values in density expression is as follows:

  $\begin{array}{c}\text{13}\text{.6}\text{g}/{\text{cm}}^{\text{3}}\text{=}\frac{\text{50}\text{.0g}}{\text{Volume}}\\ \text{Volume=}\frac{\text{50}\text{.0g}}{13.6\text{g}/{\text{cm}}^{\text{3}}}\\ \text{=3}{\text{.68cm}}^{\text{3}}\\ =3.68\text{mL}\end{array}$

Thus volume of mercury is 3.68mL.

(c)

Interpretation Introduction

Interpretation: The volume of 50.0 g benzene to be calculated.

Concept introduction: Amount of matter present in a substance in given volume is density. That is, density is ratio of mass of an object to its volume or mass per unit volume.

Answer to Problem 61A

Volume of benzene is 56.8mL.

Explanation of Solution

Expression for density is as follows:

  $\text{Density=}\frac{\text{Mass}}{\text{Volume}}$

Given, density of benzene is $\text{0}\text{.880}\text{g}/{\text{cm}}^{\text{3}}$ and mass is 50.0g. Put these values in density expression is as follows:

  $\begin{array}{c}\text{0}\text{.880}\text{g}/{\text{cm}}^{\text{3}}\text{=}\frac{\text{50}\text{.0g}}{\text{Volume}}\\ \text{Volume=}\frac{\text{50}\text{.0g}}{0.880\text{g}/{\text{cm}}^{\text{3}}}\\ \text{=56}{\text{.8cm}}^{\text{3}}\\ =56.8\text{mL}\end{array}$

Thus volume of benzene is 56.8mL.

(d)

Interpretation Introduction

Interpretation: The volume of 50.0 g silver to be calculated.

Concept introduction: Amount of matter present in a substance in given volume is density. That is, density is ratio of mass of an object to its volume or mass per unit volume.

Answer to Problem 61A

Volume of silver is 4.76mL.

Explanation of Solution

Expression for density is as follows:

  $\text{Density=}\frac{\text{Mass}}{\text{Volume}}$

Given, density of silver is $\text{10}\text{.5}\text{g}/{\text{cm}}^{\text{3}}$ and mass is 50.0g. Put these values in density expression is as follows:

  $\begin{array}{c}\text{10}\text{.5}\text{g}/{\text{cm}}^{\text{3}}\text{=}\frac{\text{50}\text{.0g}}{\text{Volume}}\\ \text{Volume=}\frac{\text{50}\text{.0g}}{10.5\text{g}/{\text{cm}}^{\text{3}}}\\ \text{=4}{\text{.76cm}}^{\text{3}}\\ =4.76\text{mL}\end{array}$

Thus volume of silver is 4.76mL.