Chapter 9, Problem 102QP

(a)

Interpretation Introduction

Interpretation:

The determination of the molar mass of a gas if its density is $1.785{\text{ g L}}^{-1}$ at $27°\text{C}$ and $745\text{ torr}$ .

Explanation of Solution

The ideal gas equation is an equation that determines the relation between the pressure $\left(P\right)$ , temperature $\left(T\right)$ , volume $\left(V\right)$ and number of moles $\left(n\right)$ of a gas. It is derived from Boyle’s law, Charles’s law, Gay-Lussac’s law, and Avogadro’s law. It states that pressure and volume are directly proportional to the number of moles and temperature of a gas.

$\begin{array}{l}PV\propto nT\\ PV=nRT\end{array}$ ….. (1)

The value of R in ${\text{L atm mol}}^{-1}{\text{ K}}^{-1}$ is $0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}$ .

The molar mass of a gas at STP can be determined by using the value of its density. The density $\left(\rho \right)$ of a gas is equal to the amount or mass $\left(m\right)$ of the gas per unit volume $\left(V\right)$ .

$\rho =\frac{m}{V}$ …… (2)

So, the molar mass of the gas can be determined by using molar volume of the gas, that is, assume $1\text{ mol}$ of gas is present at a temperature of $27°\text{C}$ and pressure of $745\text{ torr}$ .

The conversion factor for temperature from Celsius to Kelvin is as follows.

$\text{K}=273.15+°\text{C}$

For $27°\text{C}$ ,

$\begin{array}{c}\text{K}=27°\text{C+273}\text{.15}\\ =298.15\text{ K}\end{array}$

The conversion of pressure from $\text{torr}$ to $\text{atm}$ is as follows.

$\begin{array}{c}760\text{ torr}=1\text{ atm}\\ \text{1}=\frac{1\text{ atm}}{760\text{ torr}}\end{array}$

For $745\text{ torr}$ ,

$\begin{array}{c}\text{745 torr}=\frac{1\text{ atm}}{760\text{ torr}}\times \text{745 torr}\\ =0.98\text{ atm}\end{array}$

Substitute the values in equation $\left(1\right)$ to find the volume of the gas.

$\begin{array}{c}0.98\text{ atm}\times V=1\text{ mol}\times \left(0.08206{\text{ L atm mol}}^{-1}{\text{K}}^{-1}\right)\times 298.15\text{ K}\\ =\frac{1\text{ mol}\times \left(\begin{array}{l}0.08206\text{ L }\\ {\text{atm mol}}^{-1}{\text{K}}^{-1}\end{array}\right)\times 298.15\text{ K}}{0.98\text{ atm}}\\ =25.1\text{ L}\end{array}$

The volume of the gas is, therefore, $25.1\text{ L}$ and its density is $2.436{\text{ g L}}^{-1}$ . Substitute the values in equation $\left(2\right)$ to find the mass of the gas.

$\begin{array}{c}2.436{\text{ g L}}^{-1}=\frac{mass}{25.1\text{ L}}\\ mass=2.436{\text{ g L}}^{-1}\times 25.1\text{ L}\\ =61.2\text{ g}\end{array}$

The number of moles of the gas is equal to the ratio of the mass of the gas to the molar mass of that gas.

$n=\frac{mass}{molarmass}$

At STP, the number of moles of the gas is $1\text{ mol}$ and the mass of the gas is $61.2\text{ g}$ . So, the molar mass of the gas can be determined by substituting the values in the above expression.

$\begin{array}{c}1\text{ mol}=\frac{61.2\text{ g}}{molarmass}\\ molarmass=61.2{\text{ g mol}}^{-1}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The determination of the molar mass of a gas if its density is $0.842{\text{ g L}}^{-1}$ at $27°\text{C}$ and $745\text{ torr}$ .

Explanation of Solution

Since the values of pressure and temperature are the same as those of the previous part, the volume of the gas is the same, that is, $25.1\text{L}$ .

The volume of the gas is, therefore, $25.1\text{ L}$ and its density is $2.436{\text{ g L}}^{-1}$ . Substitute the values in equation $\left(2\right)$ to find the mass of the gas.

$\begin{array}{c}0.842{\text{ g L}}^{-1}=\frac{mass}{25.1\text{ L}}\\ mass=0.842{\text{ g L}}^{-1}\times 25.1\text{ L}\\ =21.1\text{ g}\end{array}$

The number of moles of the gas is equal to the ratio of the mass of the gas to the molar mass of that gas.

$n=\frac{mass}{molarmass}$

At STP, the number of moles of the gas is $1\text{ mol}$ and the mass of the gas is $21.2\text{ g}$ . So, the molar mass of the gas can be determined by substituting the values in the above expression.

$\begin{array}{c}1\text{ mol}=\frac{21.1\text{ g}}{molarmass}\\ molarmass=21.1{\text{ g mol}}^{-1}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The determination of the molar mass of a gas if its density is $1.325{\text{ g L}}^{-1}$ at $27°\text{C}$ and $745\text{ torr}$ .

Explanation of Solution

Since the values of pressure and temperature are the same as those of the previous part, the volume of the gas is the same, that is, $25.1\text{L}$ .

The volume of the gas is, therefore, $25.1\text{ L}$ and its density is $1.325{\text{ g L}}^{-1}$ . Substitute the values in equation $\left(2\right)$ to find the mass of the gas.

$\begin{array}{c}1.325{\text{ g L}}^{-1}=\frac{mass}{25.1\text{ L}}\\ mass=1.325{\text{ g L}}^{-1}\times 25.1\text{ L}\\ =33.3\text{ g}\end{array}$

The number of moles of the gas is equal to the ratio of the mass of the gas to the molar mass of that gas.

$n=\frac{mass}{molarmass}$

At STP, the number of moles of the gas is $1\text{ mol}$ and the mass of the gas is $33.3\text{ g}$ . So, the molar mass of the gas can be determined by substituting the values in the above expression.

$\begin{array}{c}1\text{ mol}=\frac{33.3\text{ g}}{molarmass}\\ molarmass=33.3{\text{ g mol}}^{-1}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The determination of the molar mass of a gas if its density is $3.450{\text{ g L}}^{-1}$ at $27°\text{C}$ and $745\text{ torr}$ .

Explanation of Solution

Since the values of pressure and temperature are the same as those of the previous part, the volume of the gas is the same, that is, $25.1\text{L}$ .

The volume of the gas is, therefore, $25.1\text{ L}$ and its density is $3.450{\text{ g L}}^{-1}$ . Substitute the values in equation $\left(2\right)$ to find the mass of the gas.

$\begin{array}{c}3.450{\text{ g L}}^{-1}=\frac{mass}{25.1\text{ L}}\\ mass=3.450{\text{ g L}}^{-1}\times 25.1\text{ L}\\ \text{=86}\text{.7 g}\end{array}$

The number of moles of the gas is equal to the ratio of the mass of the gas to the molar mass of that gas.

$n=\frac{mass}{molarmass}$

At STP, the number of moles of the gas is $1\text{ mol}$ and the mass of the gas is $86.7\text{ g}$ . So, the molar mass of the gas can be determined by substituting the values in the above expression.

$\begin{array}{c}1\text{ mol}=\frac{86.7\text{ g}}{molarmass}\\ molarmass=86.7{\text{ g mol}}^{-1}\end{array}$

(e)

Interpretation Introduction

Interpretation:

The determination of the molar mass of a gas if its density is $1.706{\text{ g L}}^{-1}$ at $27°\text{C}$ and $745\text{ torr}$ .

Explanation of Solution

Since the values of pressure and temperature are the same as those of the above part, the volume of the gas is the same, that is, $25.1\text{L}$ .

The volume of the gas is, therefore, $25.1\text{ L}$ and its density is $1.706{\text{ g L}}^{-1}$ . Substitute the values in equation $\left(2\right)$ to find the mass of the gas.

$\begin{array}{c}1.706{\text{ g L}}^{-1}=\frac{mass}{25.1\text{ L}}\\ mass=1.706{\text{ g L}}^{-1}\times 25.1\text{ L}\\ =42.9\text{ g}\end{array}$

The number of moles of the gas is equal to the ratio of the mass of the gas to the molar mass of that gas.

$n=\frac{mass}{molarmass}$

At STP, the number of moles of the gas is $1\text{ mol}$ and the mass of the gas is $42.9\text{ g}$ . So, the molar mass of the gas can be determined by substituting the values in the above expression.

$\begin{array}{c}1\text{ mol}=\frac{42.9\text{ g}}{molarmass}\\ molarmass=42.9{\text{ g mol}}^{-1}\end{array}$