Chapter 9, Problem 101QP

(a)

Interpretation Introduction

Interpretation:

The determination of the molar mass of a gas at STP if its density is $1.785{\text{ g L}}^{-1}$ .

Explanation of Solution

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 $\left(V\right)$ .

$\rho =\frac{m}{V}$ …… (1)

The density of the gas is $1.785{\text{ g L}}^{-1}$ and, at STP, the volume of the gas is $22.414\text{ L}$ .

Substitute the given values in the above expression:

$\begin{array}{c}1.785{\text{ g L}}^{-1}=\frac{m}{22.414\text{ L}}\\ m=1.785{\text{ g L}}^{-1}\times 22.414\text{ L}\\ =40.01\text{ g}\end{array}$

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 a gas is $1\text{ mol}$ and the mass of the gas is $40.01\text{ g}$ . So, the molar mass of the gas can be determined by substituting the values in the above expression.

$\begin{array}{c}1\text{ mol}=\frac{40.01\text{ g}}{molarmass}\\ molarmass=40.01{\text{ g mol}}^{-1}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The determination of the molar mass of a gas at STP if its density is $1.340{\text{ g L}}^{-1}$ .

Explanation of Solution

The density of the gas is $1.340{\text{ g L}}^{-1}$ and, at STP, the volume of the gas is $22.414\text{ L}$ . Determine the mass of the gas by substituting the values in equation $\left(1\right)$ .

$\begin{array}{c}1.340{\text{ g L}}^{-1}=\frac{m}{22.414\text{ L}}\\ m=1.340{\text{ g L}}^{-1}\times 22.414\text{ L}\\ =30.03\text{ g}\end{array}$

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 $30.03\text{ g}$ . So, the molar mass of gas can be determined by substituting the values in the above expression.

$\begin{array}{c}1\text{ mol}=\frac{30.03\text{ g}}{molarmass}\\ molarmass=30.03{\text{ g mol}}^{-1}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The determination of the molar mass of a gas at STP if its density is $2.052{\text{ g L}}^{-1}$ .

Explanation of Solution

The density of the gas is $2.052{\text{ g L}}^{-1}$ and, at STP, the volume of the gas is $22.414\text{ L}$ . Determine the mass of the gas by substituting the values in equation $\left(1\right)$ .

$\begin{array}{c}2.052{\text{ g L}}^{-1}=\frac{m}{22.414\text{ L}}\\ m=2.052{\text{ g L}}^{-1}\times 22.414\text{ L}\\ =45.99\text{ g}\end{array}$

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 $45.99\text{ g}$ . So, the molar mass of the gas can be determined by substituting the values in the above expression.

$\begin{array}{c}1\text{ mol}=\frac{45.99\text{ g}}{molarmass}\\ molarmass=45.99{\text{ g mol}}^{-1}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The determination of the molar mass of a gas at STP if its density is $0.905{\text{ g L}}^{-1}$ .

Explanation of Solution

The density of the gas is $0.905{\text{ g L}}^{-1}$ and, at STP, the volume of the gas is $22.414\text{ L}$ . Determine the mass of the gas by substituting the values in equation $\left(1\right)$ .

$\begin{array}{c}0.905{\text{ g L}}^{-1}=\frac{m}{22.414\text{ L}}\\ m=0.905{\text{ g L}}^{-1}\times 22.414\text{ L}\\ =20.3\text{ g}\end{array}$

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 $20.03\text{ g}$ . So, the molar mass of the gas can be determined by substituting the values in the above expression.

$\begin{array}{c}1\text{ mol}=\frac{20.3\text{ g}}{molarmass}\\ molarmass=20.3{\text{ g mol}}^{-1}\end{array}$

(e)

Interpretation Introduction

Interpretation:

The determination of the molar mass of a gas at STP if its density is $0.714{\text{ g L}}^{-1}$ .

Explanation of Solution

The density of the gas is $0.714{\text{ g L}}^{-1}$ and, at STP, the volume of the gas is $22.414\text{ L}$ . Determine the mass of the gas by substituting the values in equation $\left(1\right)$ .

$\begin{array}{c}0.714{\text{ g L}}^{-1}=\frac{m}{22.414\text{ L}}\\ m=0.714{\text{ g L}}^{-1}\times 22.414\text{ L}\\ =16\text{ g}\end{array}$

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 $16\text{ g}$ . So, the molar mass of the gas can be determined by substituting the values in the above expression.

$\begin{array}{c}1\text{ mol}=\frac{16\text{ g}}{molarmass}\\ molarmass=16{\text{ g mol}}^{-1}\end{array}$