Chapter 3, Problem 18CR

a. Given that 1100. mL of ethyl alcohol weighs 78.5 g, calculate the density of ethyl alcohol.

b. What volume would 1.59 kg of ethyl alcohol occupy?

c. What is the mass of 1.35 L of ethyl alcohol?

d. Pure aluminum metal has a density of 2.70 g/cm³. Calculate the volume of 25.2 g of pure aluminum.

e. What will a rectangular block of pure aluminum having dimensions of 12.0 cm × 2.5 cm × 2.5 cm weigh?

Interpretation Introduction

(a)

Interpretation:

The density of ethyl alcohol should be calculated.

Concept Introduction:

Density is mass per unit volume, the ratio of the mass of an object to its volume.

$\text{D}\text{e}\text{n}\text{s}\text{i}\text{t}\text{y}=\frac{\text{m}\text{a}\text{s}\text{s}}{\text{v}\text{o}\text{l}\text{u}\text{m}\text{e}}$.

Answer to Problem 18CR

$0.785\text{g}$.

Explanation of Solution

$\text{D}\text{e}\text{n}\text{s}\text{i}\text{t}\text{y}=\frac{\text{m}\text{a}\text{s}\text{s}}{\text{v}\text{o}\text{l}\text{u}\text{m}\text{e}}$

$\text{D}\text{e}\text{n}\text{s}\text{i}\text{t}\text{y}=\frac{78.5\text{g}}{100.\text{m}\text{L}}$ = $0.785\text{g}/\text{m}\text{L}$.

Interpretation Introduction

(b)

Interpretation:

Volume of $1.59\text{k}\text{g}$ of ethyl alcohol occupied should be calculated.

Concept Introduction:

Density is mass per unit volume, the ratio of the mass of an object to its volume.

$\text{D}\text{e}\text{n}\text{s}\text{i}\text{t}\text{y}=\frac{\text{m}\text{a}\text{s}\text{s}}{\text{v}\text{o}\text{l}\text{u}\text{m}\text{e}}$.

Answer to Problem 18CR

$2.03\text{L}$.

Explanation of Solution

$\text{D}\text{e}\text{n}\text{s}\text{i}\text{t}\text{y}=\frac{\text{m}\text{a}\text{s}\text{s}}{\text{v}\text{o}\text{l}\text{u}\text{m}\text{e}}$

$\text{V}\text{o}\text{l}\text{u}\text{m}\text{e}=\frac{\text{m}\text{a}\text{s}\text{s}}{\text{d}\text{e}\text{n}\text{s}\text{i}\text{t}\text{y}}$

$\text{V}\text{o}\text{l}\text{u}\text{m}\text{e}=\frac{1590\text{g}}{0.785\text{g}/\text{m}\text{L}}$ = $2025.4\text{m}\text{L}$

= $2.03\text{L}$.

Interpretation Introduction

(c)

Interpretation:

Mass of $1.35\text{L}$ of ethyl alcohol should be calculated.

Concept Introduction:

Density is mass per unit volume, the ratio of the mass of an object to its volume.

$\text{D}\text{e}\text{n}\text{s}\text{i}\text{t}\text{y}=\frac{\text{m}\text{a}\text{s}\text{s}}{\text{v}\text{o}\text{l}\text{u}\text{m}\text{e}}$.

Answer to Problem 18CR

$1.06\text{k}\text{g}$.

Explanation of Solution

$\text{D}\text{e}\text{n}\text{s}\text{i}\text{t}\text{y}=\frac{\text{m}\text{a}\text{s}\text{s}}{\text{v}\text{o}\text{l}\text{u}\text{m}\text{e}}$

$\text{M}\text{a}\text{s}\text{s}=\text{D}\text{e}\text{n}\text{s}\text{i}\text{t}\text{y}\times \text{V}\text{o}\text{l}\text{u}\text{m}\text{e}$

$\begin{array}{l} =0.785\text{g}/\text{m}\text{L}\times 1350\text{m}\text{L}\hfill \\ =1059.75\text{g}\hfill \\ =1.06\text{k}\text{g}\hfill \end{array}$.

Interpretation Introduction

(d)

Interpretation:

Volume of $25.2\text{g}$ of pure aluminum should be calculated.

Concept Introduction:

Density is mass per unit volume, the ratio of the mass of an object to its volume.

$\text{D}\text{e}\text{n}\text{s}\text{i}\text{t}\text{y}=\frac{\text{m}\text{a}\text{s}\text{s}}{\text{v}\text{o}\text{l}\text{u}\text{m}\text{e}}$.

Answer to Problem 18CR

$9.33\text{c}{\text{m}}^3$.

Explanation of Solution

$\text{D}\text{e}\text{n}\text{s}\text{i}\text{t}\text{y}=\frac{\text{m}\text{a}\text{s}\text{s}}{\text{v}\text{o}\text{l}\text{u}\text{m}\text{e}}$

$\text{V}\text{o}\text{l}\text{u}\text{m}\text{e}=\frac{\text{m}\text{a}\text{s}\text{s}}{\text{d}\text{e}\text{n}\text{s}\text{i}\text{t}\text{y}}$

$=\frac{25.2\text{g}}{2.70\text{g}/\text{c}{\text{m}}^3}$ = $9.33\text{c}{\text{m}}^3$.

Interpretation Introduction

(e)

Interpretation:

Weight of rectangular block of pure aluminum given should be calculated.

Concept Introduction:

Density is mass per unit volume, the ratio of the mass of an object to its volume.

$\text{D}\text{e}\text{n}\text{s}\text{i}\text{t}\text{y}=\frac{\text{m}\text{a}\text{s}\text{s}}{\text{v}\text{o}\text{l}\text{u}\text{m}\text{e}}$.

Answer to Problem 18CR

$0.20\text{k}\text{g}$.

Explanation of Solution

$\text{V}\text{o}\text{l}\text{u}\text{m}\text{e}=\text{H}\text{e}\text{i}\text{g}\text{h}\text{t}\times \text{L}\text{e}\text{n}\text{g}\text{t}\text{h}\times \text{W}\text{i}\text{d}\text{t}\text{h}$

$\text{D}\text{e}\text{n}\text{s}\text{i}\text{t}\text{y}=\frac{\text{m}\text{a}\text{s}\text{s}}{\text{v}\text{o}\text{l}\text{u}\text{m}\text{e}}$

$\text{M}\text{a}\text{s}\text{s}=\text{D}\text{e}\text{n}\text{s}\text{i}\text{t}\text{y}\times \text{V}\text{o}\text{l}\text{u}\text{m}\text{e}$

$\begin{array}{l} =2.70\text{g}/\text{c}{\text{m}}^3\times 12.0\text{c}\text{m}\times 2.5\text{c}\text{m}\times 2.5\text{c}\text{m}\hfill \\ =202.5\text{g}\hfill \\ =0.20\text{k}\text{g}\hfill \end{array}$.