Chapter 8, Problem 26E

Interpretation Introduction

(a)

Interpretation:

The density of xenon, $\text{Xe}$ at STP is to be calculated.

Concept introduction:

The density of a substance is defined as the ratio of mass and volume. It is a characteristic property of a substance that relates the mass of that substance with the space it occupies. The SI unit of density is $\text{kg}/{\text{m}}^3$.

Answer to Problem 26E

The density of xenon, $\text{Xe}$ at STP is $5.86\text{g}/\text{L}$.

Explanation of Solution

The density of xenon, $\text{Xe}$ at STP is calculated by the formula shown below.

$\text{Density}=M_{\text{Xe}}\times \frac{\text{1}\text{mole}}{\text{22}\text{.4}\text{L}}$ …(1)

Where,

• $M_{\text{Xe}}$ is molar mass of xenon.

Molar mass of the xenon $\left(M_{\text{Xe}}\right)$ is $\text{131}\text{.293}\text{g}/\text{mol}$.

Substitute the value of $M_{\text{Xe}}$ in equation (1).

$\begin{array}{rll}\text{Density}& =& M_{\text{Xe}}\times \frac{\text{1}\text{mole}}{\text{22}\text{.4}\text{L}}\\ & =& \text{131}\text{.293}\text{g}/\text{mol}\times \frac{\text{1}\text{mole}}{\text{22}\text{.4}\text{L}}\\ & =& \frac{\text{131}\text{.293}}{\text{22}\text{.4}}\text{g}/\text{L}\\ & =& 5.86\text{g}/\text{L}\end{array}$

Therefore, the density of xenon, $\text{Xe}$ is $5.86\text{g}/\text{L}$.

Conclusion

The density of xenon, $\text{Xe}$ at STP is $5.86\text{g}/\text{L}$.

Interpretation Introduction

(b)

Interpretation:

The density of chlorine, ${\text{Cl}}_{\text{2}}$ at STP is to be calculated.

Concept introduction:

The density of a substance is defined as the ratio of mass and volume. It is a characteristic property of a substance that relates the mass of that substance with the space it occupies. The SI unit of density is $\text{kg}/{\text{m}}^3$.

Answer to Problem 26E

The density of chlorine, ${\text{Cl}}_{\text{2}}$ at STP is $3.165\text{g}/\text{L}$.

Explanation of Solution

The density of chlorine, ${\text{Cl}}_{\text{2}}$ at STP is calculated by the formula shown below.

$\text{Density}=M_{{\text{Cl}}_2}\times \frac{\text{1}\text{mole}}{\text{22}\text{.4}\text{L}}$ …(1)

Where,

• $M_{{\text{Cl}}_2}$ is molar mass of chlorine.

Molar mass of the chlorine $\left(M_{{\text{Cl}}_2}\right)$ is $\text{70}\text{.906}\text{g}/\text{mol}$.

Substitute the value of $M_{{\text{Cl}}_2}$ in equation (1).

$\begin{array}{rll}\text{Density}& =& M_{{\text{Cl}}_2}\times \frac{\text{1}\text{mole}}{\text{22}\text{.4}\text{L}}\\ & =& \text{70}\text{.906}\text{g}/\text{mol}\times \frac{\text{1}\text{mole}}{\text{22}\text{.4}\text{L}}\\ & =& \frac{\text{70}\text{.906}}{\text{22}\text{.4}}\text{g}/\text{L}\\ & =& 3.165\text{g}/\text{L}\end{array}$

Therefore, the density of chlorine, ${\text{Cl}}_{\text{2}}$ is $3.165\text{g}/\text{L}$.

Conclusion

The density of chlorine, ${\text{Cl}}_{\text{2}}$ at STP is $3.165\text{g}/\text{L}$.

Interpretation Introduction

(c)

Interpretation:

The density of ethane, ${\text{C}}_2{\text{H}}_{\text{6}}$ at STP is to be calculated.

Concept introduction:

The density of a substance is defined as the ratio of mass and volume. It is a characteristic property of a substance that relates the mass of that substance with the space it occupies. The SI unit of density is $\text{kg}/{\text{m}}^3$.

Answer to Problem 26E

The density of ethane, ${\text{C}}_2{\text{H}}_{\text{6}}$ at STP is $1.34\text{g}/\text{L}$.

Explanation of Solution

The density of ethane ${\text{C}}_2{\text{H}}_{\text{6}}$ at STP is calculated by the formula shown below.

$\text{Density}=M_{{\text{C}}_2{\text{H}}_{\text{6}}}\times \frac{\text{1}\text{mole}}{\text{22}\text{.4}\text{L}}$ …(1)

Where,

• $M_{{\text{C}}_2{\text{H}}_{\text{6}}}$ is molar mass of ethane.

Molar mass of the ethane $\left(M_{{\text{C}}_2{\text{H}}_{\text{6}}}\right)$ is $\text{30}\text{.07}\text{g}/\text{mol}$.

Substitute the value of $M_{{\text{C}}_2{\text{H}}_{\text{6}}}$ in equation (1).

$\begin{array}{rll}\text{Density}& =& M_{{\text{C}}_2{\text{H}}_{\text{6}}}\times \frac{\text{1}\text{mole}}{\text{22}\text{.4}\text{L}}\\ & =& \text{30}\text{.07}\text{g}/\text{mol}\times \frac{\text{1}\text{mole}}{\text{22}\text{.4}\text{L}}\\ & =& \frac{\text{30}\text{.07}}{\text{22}\text{.4}}\text{g}/\text{L}\\ & =& 1.34\text{g}/\text{L}\end{array}$

Therefore, the density of ethane, ${\text{C}}_2{\text{H}}_{\text{6}}$ is $1.34\text{g}/\text{L}$.

Conclusion

The density of ethane, ${\text{C}}_2{\text{H}}_{\text{6}}$ at STP is $1.34\text{g}/\text{L}$.

Interpretation Introduction

(d)

Interpretation:

The density of butane, ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}$ at STP is to be calculated.

Concept introduction:

The density of a substance is defined as the ratio of mass and volume. It is a characteristic property of a substance that relates the mass of that substance with the space it occupies. The SI unit of density is $\text{kg}/{\text{m}}^3$.

Answer to Problem 26E

The density of butane, ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$ at STP is $2.59\text{g}/\text{L}$.

Explanation of Solution

The density of butane ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$ at STP is calculated by the formula shown below.

$\text{Density}=M_{{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}}\times \frac{\text{1}\text{mole}}{\text{22}\text{.4}\text{L}}$ …(1)

Where,

• $M_{{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}}$ is molar mass of butane.

Molar mass of the butane $\left(M_{{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}}\right)$ is $\text{58}\text{.12}\text{g}/\text{mol}$.

Substitute the value of $M_{{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}}$ in equation (1).

$\begin{array}{rll}\text{Density}& =& M_{{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}}\times \frac{\text{1}\text{mole}}{\text{22}\text{.4}\text{L}}\\ & =& \text{58}\text{.12}\text{g}/\text{mol}\times \frac{\text{1}\text{mole}}{\text{22}\text{.4}\text{L}}\\ & =& \frac{\text{58}\text{.12}}{\text{22}\text{.4}}\text{g}/\text{L}\\ & =& 2.59\text{g}/\text{L}\end{array}$

Therefore, the density of butane, ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$ is $2.59\text{g}/\text{L}$.

Conclusion

The density of butane, ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$ at STP is $2.59\text{g}/\text{L}$.