Chapter 8, Problem 29E

Interpretation Introduction

Interpretation:

The given table is to be completed.

Concept introduction:

One mole of every gas occupies the same volume at STP. This value is equal to $\text{22}\text{.4}\text{L}$ and is known as the molar volume of the gas. The number of particles present in $\text{1}\text{mol}$ of any gas is equal to $6.022\times {10}^{23}$ and this value is known as the Avogadro’s number.

Answer to Problem 29E

The given table is completed as shown below.

Gas	Molecules	Mass	STP Volume
nitrogen, ${\text{N}}_{\text{2}}$	$6.022\times {10}^{23}$	$28.02\text{g}$	$\text{22}\text{.4}\text{L}$
oxygen, ${\text{O}}_{\text{2}}$	$6.022\times {10}^{23}$	$32.0\text{g}$	$\text{22}\text{.4}\text{L}$
nitrogen dioxide,	$6.022\times {10}^{23}$	$46.01\text{g}$	$\text{22}\text{.4}\text{L}$
${\text{NO}}_{\text{2}}$

Explanation of Solution

It is given that the number of moles of each gas ${\text{N}}_{\text{2}}$, ${\text{O}}_{\text{2}}$ and ${\text{NO}}_{\text{2}}$ is equal to $\text{1}\text{mol}$. The number of molecules present in $\text{1}\text{mol}$ of a gas is equal to $6.022\times {10}^{23}$. Therefore, the number of molecules of gases ${\text{N}}_{\text{2}}$, ${\text{O}}_{\text{2}}$ and ${\text{NO}}_{\text{2}}$ is equal to $6.022\times {10}^{23}$. At STP, one mole of every gas occupies a volume equal to $\text{22}\text{.4}\text{L}$ which is known as the molar volume of the gas. Therefore, the volume of the gases ${\text{N}}_{\text{2}}$, ${\text{O}}_{\text{2}}$ and ${\text{NO}}_{\text{2}}$ is equal to $\text{22}\text{.4}\text{L}$.

The mass of ${\text{N}}_{\text{2}}$ molecule is calculated as shown below.

$\text{Mass}\text{of}{\text{N}}_{\text{2}}=\text{Number}\text{of}\text{atoms}\text{of}\text{N}\times \text{Mass}\text{of}\text{1}\text{N}\text{atom}$

Substitute the values of number of atoms of $\text{N}$ and mass of $\text{1}\text{N}$ atom in the above expression.

$\begin{array}{rll}\text{Mass}\text{of}{\text{N}}_{\text{2}}& =& \text{Number}\text{of}\text{atoms}\text{of}\text{N}\times \text{Mass}\text{of}\text{1}\text{N}\text{atom}\\ & =& \text{2}\times 14.01\text{g}/\text{mol}\\ & =& 28.02\text{g}/\text{mol}\end{array}$

Therefore, the mass of ${\text{N}}_{\text{2}}$ gas is $28.02\text{g}/\text{mol}$.

The mass of ${\text{O}}_{\text{2}}$ molecule is calculated as shown below.

$\text{Mass}\text{of}{\text{O}}_{\text{2}}=\text{Number}\text{of}\text{atoms}\text{of}\text{O}\times \text{Mass}\text{of}\text{1}\text{O}\text{atom}$

Substitute the values of number of atoms of $\text{O}$ and mass of $\text{1}\text{O}$ atom in the above expression.

$\begin{array}{rll}\text{Mass}\text{of}{\text{O}}_{\text{2}}& =& \text{Number}\text{of}\text{atoms}\text{of}\text{O}\times \text{Mass}\text{of}\text{1}\text{O}\text{atom}\\ & =& \text{2}\times 16.0\text{g}/\text{mol}\\ & =& 32.0\text{g}/\text{mol}\end{array}$

Therefore, the mass of ${\text{O}}_{\text{2}}$ gas is $32.0\text{g}/\text{mol}$.

The mass of ${\text{NO}}_{\text{2}}$ molecule is calculated as shown below.

$\text{Mass}\text{of}{\text{NO}}_{\text{2}}=\left(\begin{array}{c}\left(\text{Number}\text{of}\text{atoms}\text{of}\text{N}\times \text{Mass}\text{of}\text{1}\text{N}\text{atom}\right)+\\ \left(\text{Number}\text{of}\text{atoms}\text{of}\text{O}\times \text{Mass}\text{of}\text{1}\text{O}\text{atom}\right)\end{array}\right)$

Substitute the values of number of atoms and mass of $\text{1}$ atom of nitrogen and oxygen in the above expression.

$\begin{array}{rll}\text{Mass}\text{of}{\text{NO}}_{\text{2}}& =& \left(\begin{array}{c}\left(\text{Number}\text{of}\text{atoms}\text{of}\text{N}\times \text{Mass}\text{of}\text{1}\text{N}\text{atom}\right)+\\ \left(\text{Number}\text{of}\text{atoms}\text{of}\text{O}\times \text{Mass}\text{of}\text{1}\text{O}\text{atom}\right)\end{array}\right)\\ & =& \left(\text{1}\times 14.01\text{g}/\text{mol}\right)+\left(\text{2}\times 16.0\text{g}/\text{mol}\right)\\ & =& 14.01\text{g}/\text{mol}+32.0\text{g}/\text{mol}\\ & =& 46.01\text{g}/\text{mol}\end{array}$

Therefore, the mass of ${\text{NO}}_{\text{2}}$ gas is $46.01\text{g}/\text{mol}$.

Therefore, the given table is completed as shown below.

Gas	Molecules	Mass	STP Volume
nitrogen, ${\text{N}}_{\text{2}}$	$6.022\times {10}^{23}$	$28.02\text{g}$	$\text{22}\text{.4}\text{L}$
oxygen, ${\text{O}}_{\text{2}}$	$6.022\times {10}^{23}$	$32.0\text{g}$	$\text{22}\text{.4}\text{L}$
nitrogen dioxide,	$6.022\times {10}^{23}$	$46.01\text{g}$	$\text{22}\text{.4}\text{L}$
${\text{NO}}_{\text{2}}$

Conclusion

The given table is completed as shown below.

Gas	Molecules	Mass	STP Volume
nitrogen, ${\text{N}}_{\text{2}}$	$6.022\times {10}^{23}$	$28.02\text{g}$	$\text{22}\text{.4}\text{L}$
oxygen, ${\text{O}}_{\text{2}}$	$6.022\times {10}^{23}$	$32.0\text{g}$	$\text{22}\text{.4}\text{L}$
nitrogen dioxide,	$6.022\times {10}^{23}$	$46.01\text{g}$	$\text{22}\text{.4}\text{L}$
${\text{NO}}_{\text{2}}$