Chapter 1, Problem 77E

In e-ach of the following pairs, which has the greater mass? (See Table 1.5.)

1. a. 1.0 kg of feathers or 1.0 kg of lead
2. b. 1.0 mL of mercury or 1.0 mL of water
3. c. 19.3 mL of water or 1.00 mL of gold
4. d. 75 mL of copper or 1.0 L of benzene

Interpretation Introduction

Interpretation:

From the given set of pairs the substance which has greater mass has to be identified.

Concept introduction:

Density:

The relationship between mass, density and volume of the substance is given by

$\begin{array}{l}\text{d}\text{=}\frac{\text{m}}{\text{V}}\\ \text{Where,}\text{d}\text{-}\text{density}\\ \text{m}\text{-}\text{mass}\\ \text{V}\text{-}\text{volume}\end{array}$

To identify: The substance which has greater mass from the given pair

Answer to Problem 77E

Greater mass is exhibited by Lead

Explanation of Solution

The identification of the substance which has greater mass from the given pair is given below

Given data

$\begin{array}{l}\text{1}\text{.0}\text{kg}\text{of}\text{feathers}\text{and}\text{1}\text{.0}\text{kg}\text{of}\text{lead}\text{.}\\ \text{density of lead = 11}{\text{.34 g/cm}}^{\text{3}}\text{.}\end{array}$

Give the relationship between mass and density

$\begin{array}{l}\text{d}\text{=}\frac{\text{m}}{\text{V}}\\ \text{therefore,}\text{density}\text{is}\text{directly}\text{proportinal}\text{mass}\text{of}\text{the}\text{substance}\end{array}$

From the given pair, lead has the greater mass because of its higher density compare to feathers.

Interpretation Introduction

Interpretation:

From the given set of pairs the substance which has greater mass has to be identified.

Concept introduction:

Density:

The relationship between mass, density and volume of the substance is given by

$\begin{array}{l}\text{d}\text{=}\frac{\text{m}}{\text{V}}\\ \text{Where,}\text{d}\text{-}\text{density}\\ \text{m}\text{-}\text{mass}\\ \text{V}\text{-}\text{volume}\end{array}$

To identify: The substance which has greater mass from the given pair

Answer to Problem 77E

Greater mass is exhibited by Mercury

Explanation of Solution

The identification of the substance which has greater mass from the given pair is given below

Given data

$\begin{array}{l}\text{1}\text{.0}\text{mL}\text{of}\text{mercury}\text{and}\text{1}\text{.0}\text{mL}\text{of}\text{water}\text{.}\\ \text{density of mercury = 13}{\text{.56 g/cm}}^{\text{3}}\text{.}\\ \text{density of water = 0}{\text{.9982 g/cm}}^{\text{3}}\text{.}\end{array}$

Give the relationship between mass and density

$\begin{array}{l}\text{d}\text{=}\frac{\text{m}}{\text{V}}\\ \text{therefore,}\text{density}\text{is}\text{directly}\text{proportinal}\text{mass}\text{of}\text{the}\text{substance}\end{array}$

Calculate the mass of mercury by the use of given data

$\begin{array}{l}\text{m}\text{=}\text{Vd}\\ \text{=}\text{1}\text{.0}{\text{cm}}^{\text{3}}\text{×}\frac{\text{13}\text{.56}\text{g}}{{\text{cm}}^{\text{3}}}\text{=}\text{13}\text{.56}\text{g}\end{array}$

Calculate the mass of water by the use of given data

$\begin{array}{l}\text{m}\text{=}\text{Vd}\\ \text{=}\text{1}{\text{cm}}^{\text{3}}\text{×}\frac{\text{0}\text{.9982}\text{g}}{{\text{cm}}^{\text{3}}}\text{=}\text{0}\text{.9982}\text{g}\end{array}$

From the given pair, mercury has the greater mass because of its higher density compare to water.

Interpretation Introduction

Interpretation:

From the given set of pairs the substance which has greater mass has to be identified.

Concept introduction:

Density:

The relationship between mass, density and volume of the substance is given by

$\begin{array}{l}\text{d}\text{=}\frac{\text{m}}{\text{V}}\\ \text{Where,}\text{d}\text{-}\text{density}\\ \text{m}\text{-}\text{mass}\\ \text{V}\text{-}\text{volume}\end{array}$

To identify: The substance which has greater mass from the given pair

Answer to Problem 77E

Greater mass is exhibited by Gold

Explanation of Solution

The identification of the substance which has greater mass from the given pair is given below

Given data

$\begin{array}{l}\text{19}\text{.3}\text{mL}\text{of}\text{water}\text{and}\text{1}\text{.0}\text{mL}\text{of}\text{gold}\text{.}\\ \text{density of gold = 19}{\text{.32 g/cm}}^{\text{3}}\text{.}\\ \text{density of water = 0}{\text{.9982 g/cm}}^{\text{3}}\text{.}\end{array}$

Give the relationship between mass and density

$\begin{array}{l}\text{d}\text{=}\frac{\text{m}}{\text{V}}\\ \text{therefore,}\text{density}\text{is}\text{directly}\text{proportinal}\text{mass}\text{of}\text{the}\text{substance}\end{array}$

Calculate the mass of gold by the use of given data

$\begin{array}{l}\text{m}\text{=}\text{Vd}\\ \text{=}\text{1}\text{.0}{\text{cm}}^{\text{3}}\text{×}\frac{\text{19}\text{.32}\text{g}}{{\text{cm}}^{\text{3}}}\text{=}\text{19}\text{.32}\text{g}\end{array}$

Calculate the mass of water by the use of given data

$\begin{array}{l}\text{m}\text{=}\text{Vd}\\ \text{=}\text{19}\text{.3}{\text{cm}}^{\text{3}}\text{×}\frac{\text{0}\text{.9982}\text{g}}{{\text{cm}}^{\text{3}}}\text{=}\text{19}\text{.27}\text{g}\end{array}$

From the given pair, gold has the greater mass because of its higher density compare to water.

Interpretation Introduction

Interpretation:

From the given set of pairs the substance which has greater mass has to be identified.

Concept introduction:

Density:

The relationship between mass, density and volume of the substance is given by

$\begin{array}{l}\text{d}\text{=}\frac{\text{m}}{\text{V}}\\ \text{Where,}\text{d}\text{-}\text{density}\\ \text{m}\text{-}\text{mass}\\ \text{V}\text{-}\text{volume}\end{array}$

To identify: The substance which has greater mass from the given pair

Answer to Problem 77E

Greater mass is exhibited by Benzene

Explanation of Solution

The identification of the substance which has greater mass from the given pair is given below

Given data

$\begin{array}{l}\text{75}\text{mL}\text{of}\text{copper}\text{and}\text{1}\text{.0}\text{L}\text{of}\text{benzene}\text{.}\\ \text{density of copper = 8}{\text{.96 g/cm}}^{\text{3}}\text{.}\\ \text{density of benzene = 0}{\text{.88 g/cm}}^{\text{3}}\text{.}\end{array}$

Calculate the mass of copper by the use of given data

$\begin{array}{l}\text{m}\text{=}\text{Vd}\\ \text{=}\text{75}{\text{cm}}^{\text{3}}\text{×}\frac{\text{8}\text{.96}\text{g}}{{\text{cm}}^{\text{3}}}\text{=}\text{672}\text{g}\end{array}$

Calculate the mass of benzene by the use of given data

$\begin{array}{l}\text{m}\text{=}\text{Vd}\\ \text{=}\text{1000}{\text{cm}}^{\text{3}}\text{×}\frac{\text{0}\text{.88}\text{g}}{{\text{cm}}^{\text{3}}}\text{=}\text{880}\text{g}\end{array}$

From the given pair, benzene has the greater mass because of its higher volume compare to copper.