Chapter 8, Problem 27E

Interpretation Introduction

(a)

Interpretation:

The molar mass of oxygen with density $\text{1}\text{.43}\text{g}/\text{L}$ 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 27E

The molar mass of oxygen with density $\text{1}\text{.43}\text{g}/\text{L}$ at STP is $32.032\text{g}/\text{mole}$.

Explanation of Solution

The density of oxygenn at STP is calculated by the formula shown below.

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

Where,

• $M_{{\text{O}}_2}$ is molar mass of oxygen.

The value of density is given as $\text{1}\text{.43}\text{g}/\text{L}$.

Substitute the value of density in equation (1).

$\begin{array}{rll}\text{Density}& =& M_{{\text{O}}_2}\times \frac{\text{1}\text{mole}}{\text{22}\text{.4}\text{L}}\\ \text{1}\text{.43}\text{g}/\text{L}& =& M_{{\text{O}}_2}\times \frac{\text{1}\text{mole}}{\text{22}\text{.4}\text{L}}\\ M_{{\text{O}}_2}& =& \text{1}\text{.43}\text{g}/\text{L}\times \frac{\text{22}\text{.4}\text{L}}{\text{1}\text{mole}}\\ & =& 32.032\text{g}/\text{mole}\end{array}$

Conclusion

The molar mass of oxygen with density $\text{1}\text{.43}\text{g}/\text{L}$ at STP is $32.032\text{g}/\text{mole}$.

Interpretation Introduction

(b)

Interpretation:

The molar mass of phosphine with density $\text{1}\text{.52}\text{g}/\text{L}$ 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 27E

The molar mass of phosphine with density $\text{1}\text{.52}\text{g}/\text{L}$ at STP, is $34.048\text{g}/\text{mole}$.

Explanation of Solution

The density of phosphine at STP is calculated by the formula shown below.

$\text{Density}=M_{{\text{PH}}_{\text{3}}}\times \frac{\text{1}\text{mole}}{\text{22}\text{.4}\text{L}}$ …(1)

Where,

• $M_{{\text{PH}}_{\text{3}}}$ is molar mass of phosphine.

The value of density is given as $\text{1}\text{.52}\text{g}/\text{L}$.

Substitute the value of density in equation (1).

$\begin{array}{rll}\text{Density}& =& M_{{\text{PH}}_{\text{3}}}\times \frac{\text{1}\text{mole}}{\text{22}\text{.4}\text{L}}\\ \text{1}\text{.52}\text{g}/\text{L}& =& M_{{\text{PH}}_{\text{3}}}\times \frac{\text{1}\text{mole}}{\text{22}\text{.4}\text{L}}\\ M_{{\text{PH}}_{\text{3}}}& =& \text{1}\text{.52}\text{g}/\text{L}\times \frac{\text{22}\text{.4}\text{L}}{\text{1}\text{mole}}\\ & =& 34.048\text{g}/\text{mole}\end{array}$

Conclusion

The molar mass of phosphine with density $\text{1}\text{.52}\text{g}/\text{L}$ at STP, is $34.048\text{g}/\text{mole}$.

Interpretation Introduction

(c)

Interpretation:

The molar mass of nitrous oxide with density $\text{1}\text{.97}\text{g}/\text{L}$ 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 27E

The molar mass of nitrous oxide with density $\text{1}\text{.97}\text{g}/\text{L}$ at STP is $44.128\text{g}/\text{mole}$.

Explanation of Solution

The density of nitrous oxide at STP is calculated by the formula shown below.

$\text{Density}=M_{{\text{N}}_{\text{2}}\text{O}}\times \frac{\text{1}\text{mole}}{\text{22}\text{.4}\text{L}}$ …(1)

Where,

• $M_{{\text{N}}_{\text{2}}\text{O}}$ is molar mass of nitrous oxide.

The value of density is given as $\text{1}\text{.97}\text{g}/\text{L}$.

Substitute the value of density in equation (1).

$\begin{array}{rll}\text{Density}& =& M_{{\text{N}}_{\text{2}}\text{O}}\times \frac{\text{1}\text{mole}}{\text{22}\text{.4}\text{L}}\\ \text{1}\text{.97}\text{g}/\text{L}& =& M_{{\text{N}}_{\text{2}}\text{O}}\times \frac{\text{1}\text{mole}}{\text{22}\text{.4}\text{L}}\\ M_{{\text{N}}_{\text{2}}\text{O}}& =& \text{1}\text{.97}\text{g}/\text{L}\times \frac{\text{22}\text{.4}\text{L}}{\text{1}\text{mole}}\\ & =& 44.128\text{g}/\text{mole}\end{array}$

Conclusion

The molar mass of nitrous oxide with density $\text{1}\text{.97}\text{g}/\text{L}$ at STP is $44.128\text{g}/\text{mole}$.

Interpretation Introduction

(d)

Interpretation:

The molar mass of Freon $-12$ with $\text{5}\text{.40}\text{g}/\text{L}$ 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 27E

The molar mass of Freon $-12$ with $\text{5}\text{.40}\text{g}/\text{L}$ at STP is $120.96\text{g}/\text{mole}$.

Explanation of Solution

The density of Freon $-12$ at STP is calculated by the formula shown below.

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

Where,

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

The value of density is given as $\text{5}\text{.40}\text{g}/\text{L}$.

Substitute the value of density in equation (1).

$\begin{array}{rll}\text{Density}& =& M_{\text{Freon}}\times \frac{\text{1}\text{mole}}{\text{22}\text{.4}\text{L}}\\ \text{5}\text{.40}\text{g}/\text{L}& =& M_{\text{Freon}}\times \frac{\text{1}\text{mole}}{\text{22}\text{.4}\text{L}}\\ M_{\text{Freon}}& =& \text{5}\text{.40}\text{g}/\text{L}\times \frac{\text{22}\text{.4}\text{L}}{\text{1}\text{mole}}\\ & =& 120.96\text{g}/\text{mole}\end{array}$

Conclusion

The molar mass of Freon $-12$ with $\text{5}\text{.40}\text{g}/\text{L}$ at STP is $120.96\text{g}/\text{mole}$.