Chapter 9, Problem 152QP

(a)

Interpretation Introduction

Interpretation:

The partial pressure of the water vapour is to be determined.

Explanation of Solution

Given Information: Table 9.2

The partial pressure of a gas in a mixture is defined as the force exerted by the molecules of that gas per unit area if it alone occupies the entire volume of the container in which the mixture is kept. It depends upon the mole fraction of the gas. The partial pressure of water vapor at different temperatures is standardized in table 9.2.

At $50.0°\text{C}$ , the partial pressure of water vapor is $92.5\text{ torr}$ .

(b)

Interpretation Introduction

Interpretation:

The partial pressure of oxygen using the ratio of ${\text{H}}_2\text{O}$ and ${\text{O}}_2$ gases in the molecular level image and Dalton’s law of partial pressure is to be calculated.

Explanation of Solution

The molecular level image shows that there are ${\text{O}}_2$ gas molecules as $5$ times of ${\text{H}}_2\text{O}$ molecules. So, the partial pressure of the ${\text{O}}_2$ gas is also $5$ times of ${\text{H}}_2\text{O}$ molecules. The partial pressure of water vapor is $92.5\text{ torr}$ . So, the partial pressure of ${\text{O}}_2$ gas is,

$\begin{array}{c}{P}_{{\text{O}}_2}=5\times 92.5\text{ torr}\\ =\text{462}\text{.5 torr}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The total pressure of the gas collected is to be determined.

Explanation of Solution

Dalton’s law of partial pressure states that in a mixture, the total pressure exerted by the mixture of gases is equal to the sum of partial pressures exerted by each individual gas.

$P_{total}=P_1+P_2+\mathrm{........}+P_n$

The partial pressure of water vapor is $92.5\text{ torr}$ and the partial pressure of ${\text{O}}_2$ gas is $\text{462}\text{.5 torr}$ . So, the total pressure of the gas collected is determined by using Dalton’s law of partial pressure.

$\begin{array}{c}{P}_{\text{total}}=P_{{\text{O}}_2}+P_{{\text{H}}_{\text{2}}\text{O}}\\ =\text{462}\text{.5 torr}+92.5\text{ torr}\\ =555\text{ torr}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The number of moles of oxygen gas collected is to be determined.

Explanation of Solution

The ideal gas equation states that the pressure $\left(P\right)$ and volume $\left(V\right)$ of the gas is directly proportional to the number of moles $\left(n\right)$ and temperature $\left(T\right)$ of the gas.

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

Here, $R$ is a universal gas constant.

The volume of the collected gas is $120.0\text{ mL}$ . The partial pressure of ${\text{O}}_2$ gas is $\text{462}\text{.5 torr}$ and temperature is $50.0°\text{C}$ .

Convert $\text{mL}$ into $\text{L}$ units as follows:

$\begin{array}{l}1\text{ L}=1000\text{ mL}\\ \text{1 mL}=\frac{1\text{ L}}{1000\text{ mL}}\end{array}$

Convert $120.0\text{ mL}$ into $\text{L}$ units as follows:

$\begin{array}{c}120\text{ mL}=\frac{1\text{ L}}{1000\text{ mL}}\times 120\text{ mL}\\ =0.12\text{ L}\end{array}$

Convert temperature units from degree Celsius to Kelvin as follows:

$T\left(°\text{C}\right)=T\left(\text{K}\right)+273.15$

Convert temperature $50.0°\text{C}$ into Kelvin as follows:

$\begin{array}{c}T=50°\text{C}+273.15\\ =323.15\text{ K}\end{array}$

Convert pressure units from $\text{torr}$ to $\text{atm}$ as follows:

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

Convert pressure $\text{462}\text{.5 torr}$ into $\text{atm}$ as follows:

$\begin{array}{c}462.5\text{ torr}=\frac{1\text{ atm}}{760\text{ torr}}\times 462.5\text{ torr}\\ =0.60855\text{ atm}\end{array}$

Substitute $P$ as $0.60855\text{ atm}$ , $V$ as $0.12\text{ L}$ , $R$ as $0.08206{\text{ L atm mol}}^{-1}{\text{ K}}^{-1}$ and $T$ as $323.15\text{ K}$ in equation (1) as follows:

$\begin{array}{c}0.60855\text{ atm}\times 0.12\text{ L}=n\times 0.08206{\text{ L atm mol}}^{-1}{\text{ K}}^{-1}\times 323.15\text{ K}\\ n=\frac{0.60855\text{ atm}\times 0.12\text{ L}}{0.08206{\text{ L atm mol}}^{-1}{\text{ K}}^{-1}\times 323.15\text{ K}}\\ =0.002754\text{ mol}\end{array}$

So, the number of moles of oxygen gas collected is $0.002754\text{ mol}$ .

(e)

Interpretation Introduction

Interpretation:

The number of moles of ${\text{KClO}}_3$ reacted is to be determined.

Explanation of Solution

The reaction is as follows:

$2{\text{KClO}}_3\left(s\right)\stackrel{\text{heat}}{\to }2\text{KCl}\left(s\right)+3{\text{O}}_2\left(g\right)$

According to the reaction,

$\begin{array}{l}2{\text{ mol KClO}}_3\text{ produces}=3{\text{ mol O}}_2\\ 1{\text{ mol O}}_2\text{ is produced by}=\frac{2{\text{ mol KClO}}_{\text{3}}}{3{\text{ mol O}}_2}\end{array}$

So, $0.002754\text{ mol}$ ${\text{O}}_2$ is produced by,

$\begin{array}{c}0.002754{\text{ mol O}}_2=\frac{2{\text{ mol KClO}}_{\text{3}}}{3{\text{ mol O}}_2}\times 0.002754{\text{ mol O}}_2\\ =0.00184{\text{ mol KClO}}_{\text{3}}\end{array}$

The number of moles of ${\text{KClO}}_3$ reacted is $0.00184\text{ mol}$ .