Chapter 9, Problem 163QP

(a)

Interpretation Introduction

Interpretation:

The gas which occupies the greatest volume at STP is to be determined.

Explanation of Solution

The number of moles of a gas $\left(n\right)$ is equal to the ratio of mass of the gas $\left(m\right)$ to the molar mass $\left(MM\right)$ of the gas.

$n=\frac{m}{MM}$ …… (1)

STP is the standard temperature and pressure. At STP, the volume of a gas occupied by $\text{1 mol}$ of gas at $1\text{ atm}$ and $273.15\text{ K}$ is $22.414\text{ L}$ . So, the volume occupied by a gas at STP can be determined by its molar volume.

$46.2{\text{ g N}}_2$

The number of moles of ${\text{N}}_2$ gas can be determined by using equation (1). The molar mass of ${\text{N}}_2$ gas is:

$\begin{array}{c}{\text{N}}_2=2\times 14{\text{ g mol}}^{-1}\\ =28{\text{ g mol}}^{-1}\end{array}$

Substitute $m$ as $46.2\text{ g}$ and $MM$ as $28{\text{ g mol}}^{-1}$ in equation (1):

$\begin{array}{c}n=\frac{46.2\text{ g}}{28{\text{ g mol}}^{-1}}\\ =1.65\text{ mol}\end{array}$

The volume occupied by $\text{1 mol}$ of ${\text{N}}_2$ gas at STP is:

$1\text{mol}=22.414\text{ L}$

The volume occupied by $\text{1}\text{.65 mol}$ of ${\text{N}}_2$ gas at STP is:

$\begin{array}{c}1.65\text{ mol}=\frac{22.414\text{ L}}{1\text{mol}}\times 1.65\text{ mol}\\ =37\text{ L}\end{array}$

Therefore, option (a) is the incorrect option.

(b)

Interpretation Introduction

Interpretation:

The gas which occupies the greatest volume at STP is to be determined.

Explanation of Solution

$\text{24}{\text{.3 g H}}_2$

The number of moles of ${\text{H}}_2$ gas can be determined by using equation (1). The molar mass of ${\text{H}}_2$ gas is:

$\begin{array}{c}{\text{H}}_2=2\times 1{\text{ g mol}}^{-1}\\ =2{\text{ g mol}}^{-1}\end{array}$

Substitute $m$ as $\text{24}\text{.3 g}$ and $MM$ as $2{\text{ g mol}}^{-1}$ in equation (1):

$\begin{array}{c}n=\frac{\text{24}\text{.3 g}}{2{\text{ g mol}}^{-1}}\\ =12.15\text{ mol}\end{array}$

The volume occupied by $\text{1 mol}$ of ${\text{H}}_2$ gas at STP is:

$1\text{mol}=22.414\text{ L}$

The volume occupied by $\text{12}\text{.15 mol}$ of ${\text{H}}_2$ gas at STP is:

$\begin{array}{c}12.15\text{ mol}=\frac{22.414\text{ L}}{1\text{mol}}\times 12.15\text{ mol}\\ =272.33\text{ L}\end{array}$

Therefore, ${\text{H}}_2$ gas is the gas which occupies the greatest volume at STP. Therefore, option (b) is the correct option.

(c)

Interpretation Introduction

Interpretation:

The gas which occupies the greatest volume at STP is to be determined.

Explanation of Solution

$\text{54}{\text{.1 g CO}}_2$

The number of moles of ${\text{CO}}_2$ gas can be determined by using equation (1). The molar mass of ${\text{CO}}_2$ gas is:

$\begin{array}{c}{\text{CO}}_2=12{\text{ g mol}}^{-1}+2\times 16{\text{ g mol}}^{-1}\\ =44{\text{ g mol}}^{-1}\end{array}$

Substitute $m$ as $\text{54}\text{.1 g}$ and $MM$ as ${\text{44 g mol}}^{-1}$ in equation (1):

$\begin{array}{c}n=\frac{\text{54}\text{.1 g}}{{\text{44 g mol}}^{-1}}\\ =1.23\text{ mol}\end{array}$

The volume occupied by $\text{1 mol}$ of ${\text{CO}}_2$ gas at STP is:

$1\text{mol}=22.414\text{ L}$

The volume occupied by $\text{1}\text{.23 mol}$ of ${\text{CO}}_2$ gas at STP is:

$\begin{array}{c}1.23\text{ mol}=\frac{22.414\text{ L}}{1\text{mol}}\times 1.23\text{ mol}\\ =27.6\text{ L}\end{array}$

Therefore, option (c) is the incorrect option.

(d)

Interpretation Introduction

Interpretation:

The gas which occupies the greatest volume at STP is to be determined.

Explanation of Solution

$\text{39}\text{.4 g Ar}$

The number of moles of $\text{Ar}$ gas can be determined by using equation (1). The molar mass of $\text{Ar}$ gas is:

$\text{Ar}=40{\text{ g mol}}^{-1}$

Substitute $m$ as $\text{39}\text{.4 g}$ and $MM$ as ${\text{40 g mol}}^{-1}$ in equation (1):

$\begin{array}{c}n=\frac{\text{39}\text{.4 g}}{{\text{40 g mol}}^{-1}}\\ =0.985\text{ mol}\end{array}$

The volume occupied by $\text{1 mol}$ of $\text{Ar}$ gas at STP is:

$1\text{mol}=22.414\text{ L}$

The volume occupied by $\text{0}\text{.985 mol}$ of $\text{Ar}$ gas at STP is:

$\begin{array}{c}\text{0}\text{.985 mol}=\frac{22.414\text{ L}}{1\text{mol}}\times 0.985\text{ mol}\\ =22.1\text{ L}\end{array}$

Therefore, option (d) is the incorrect option.

(e)

Interpretation Introduction

Interpretation:

The gas which occupies the greatest volume at STP is to be determined.

Explanation of Solution

$\text{21}{\text{.5 g NH}}_3$

The number of moles of ${\text{NH}}_3$ gas can be determined by using equation (1). The molar mass of ${\text{NH}}_3$ gas is:

$\begin{array}{c}{\text{NH}}_3=14{\text{ g mol}}^{-1}+3\times 1{\text{ g mol}}^{-1}\\ =17{\text{ g mol}}^{-1}\end{array}$

Substitute $m$ as $\text{21}\text{.5 g}$ and $MM$ as ${\text{17 g mol}}^{-1}$ in equation (1):

$\begin{array}{c}n=\frac{\text{21}\text{.5 g}}{{\text{17 g mol}}^{-1}}\\ =1.3\text{ mol}\end{array}$

The volume occupied by $\text{1 mol}$ of $\text{Ar}$ gas at STP is:

$1\text{mol}=22.414\text{ L}$

The volume occupied by $\text{1}\text{.3 mol}$ of $\text{Ar}$ gas at STP is:

$\begin{array}{c}\text{1}\text{.3 mol}=\frac{22.414\text{ L}}{1\text{mol}}\times 1.3\text{ mol}\\ =28.24\text{ L}\end{array}$

Therefore, option (e) is the incorrect option.

Arrange the volumes in the increasing order as follows:

$\begin{array}{l}272.33\text{ L}>37.0\text{ L}>28.24\text{ L}>27.6\text{ L}>22.1\text{ L}\\ \text{B}>\text{A}>\text{E}>\text{C}>\text{D}\end{array}$