Chapter 5, Problem 111E

Interpretation Introduction

Interpretation:Mixing ratio and number of molecule of benzene and tolueneshould be determined.

Concept introduction: State of any gas is determined by some of its parameters. These parameters are pressure, amount of gas, temperature and volume. Ideal gas law helps to govern state of gas with help of relationships between these gas parameters.

Expression for ideal gas equation is as follows:

  $PV=nRT$

Where,

• $P$ is pressure of gas.
• $V$ is volume of gas.
• $n$ is amount or moles of gas.
• $R$ is universal gas constant.
• $T$ is absolute temperature of gas.

Answer to Problem 111E

Mixing ratio of benzene and toluene is $9.47\times {10}^{-3}\text{ ppmv}$ and $1.37\times {10}^{-2}\text{ ppmv}$ respectively. Number of molecule of benzene and toluene per cubic centimeter is $2.31\times {10}^{11}{\text{ molecules/cm}}^3$ and $3.21\times {10}^{11}{\text{ molecules/cm}}^3$ respectively.

Explanation of Solution

Mole of benzene in $89.6\text{ ng}$ can be calculated as follows:

  $\begin{array}{c}\text{Mole}=\left(\frac{89.6\text{ ng}}{78.11\text{ g/mol}}\right)\left(\frac{{10}^{-9}\text{ g}}{1\text{ ng}}\right)\\ =1.15\times {10}^{-9}\text{ mol}\end{array}$

Molecules of benzene per cubic centimeter can be calculated as follows:

  $\begin{array}{c}{\text{Molecules of C}}_6{\text{H}}_6=\left(1.15\times {10}^{-9}\text{ mol}\right)\left(6.022\times {10}^{23}\text{ molecules/mol}\right)\\ =6.93\times {10}^{14}{\text{ C}}_6{\text{H}}_6\text{ molecules}\end{array}$

Since molecules of ${\text{C}}_6{\text{H}}_6$ present in $3.00\text{ L}$ thus number of ${\text{C}}_6{\text{H}}_6$ molecule per cubic centimeter can be calculated as follows:

  $\begin{array}{c}{\text{Molecules of C}}_6{\text{H}}_6=\left(\frac{6.93\times {10}^{14}{\text{ C}}_6{\text{H}}_6\text{ molecules}}{3.00\text{ L}}\right)\left(\frac{1\text{ L}}{{10}^3{\text{ cm}}^3}\right)\\ =2.31\times {10}^{11}{\text{ C}}_6{\text{H}}_6{\text{ molecules/cm}}^3\end{array}$

Hence benzene has $2.31\times {10}^{11}{\text{ molecules/cm}}^3$ in air sample.

Conversion of $23°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{T}\left(°\text{C}\right)+273\\ =23°\text{C}+273\\ =296\text{K}\end{array}$

Expression for ideal gas equation is as follows:

  $PV=nRT$

Where,

• $P$ is pressure of gas.
• $V$ is volume of gas.
• $n$ is amount or moles of gas.
• $R$ is universal gas constant.
• $T$ is absolute temperature of gas.

Rearrange above equation to calculate value of $P$ for volume of ${\text{C}}_6{\text{H}}_6$ .

  $V=\frac{nRT}{P}$

Value of $P$ is $\text{748 torr}$ .

Value of $n$ is $1.15\times {10}^{-9}\text{ mol}$ .

Value of $T$ is $296\text{K}$ .

Value of $R$ is $0.08206\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ .

Substitute the value in above equation.

  $\begin{array}{c}V=\frac{nRT}{P}\\ =\left(\frac{\left(1.15\times {10}^{-9}\text{ mol}\right)\left(0.08206\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(296\text{K}\right)}{\text{748 torr}}\right)\left(\frac{760\text{ torr}}{1\text{ atm}}\right)\\ =2.84\times {10}^{-8}\text{ L}\end{array}$

Expression of $\text{ppmv}$ for ${\text{C}}_6{\text{H}}_6$ is as follows:

  ${\text{ppmv of C}}_6{\text{H}}_6=\left(\frac{{\text{Volume of C}}_6{\text{H}}_6\text{ at STP}}{\text{Total volume of air at STP}}\right)\left({10}^6\right)$

Where,

• $X$ is trace compound.

Value of volume of ${\text{C}}_6{\text{H}}_6$ is $2.84\times {10}^{-8}\text{ L}$ .

Value of total volume equivalent of ${\text{C}}_6{\text{H}}_6$ is $3.00\text{ L}$ .

Substitute the value in above equation.

  $\begin{array}{c}{\text{ppmv of C}}_6{\text{H}}_6=\left(\frac{{\text{Volume of C}}_6{\text{H}}_6\text{ at STP}}{\text{Total volume of air at STP}}\right)\left({10}^6\right)\\ =\left(\frac{2.84\times {10}^{-8}\text{ L}}{3.00\text{ L}}\right)\left({10}^6\right)\\ =9.47\times {10}^{-3}\text{ ppmv}\end{array}$

Mole of toluene in $153\text{ ng}$ can be calculated as follows:

  $\begin{array}{c}\text{Mole}=\left(\frac{153\text{ ng}}{92.14\text{ g/mol}}\right)\left(\frac{{10}^{-9}\text{ g}}{1\text{ ng}}\right)\\ =1.66\times {10}^{-9}\text{ mol}\end{array}$

Molecules of toluene per cubic centimeter can be calculated as follows:

  $\begin{array}{c}{\text{Molecules of C}}_7{\text{H}}_8=\left(1.66\times {10}^{-9}\text{ mol}\right)\left(6.022\times {10}^{23}\text{ molecules/mol}\right)\\ =9.64\times {10}^{14}{\text{ C}}_7{\text{H}}_8\text{ molecules}\end{array}$

Since molecules of ${\text{C}}_7{\text{H}}_8$ present in $3.00\text{ L}$ thus number of ${\text{C}}_7{\text{H}}_8$ molecule per cubic centimeter can be calculated as follows:

  $\begin{array}{c}{\text{Molecules of C}}_7{\text{H}}_8=\left(\frac{9.64\times {10}^{14}{\text{ C}}_7{\text{H}}_8\text{ molecules}}{3.00\text{ L}}\right)\left(\frac{1\text{ L}}{{10}^3{\text{ cm}}^3}\right)\\ =3.21\times {10}^{11}{\text{ C}}_7{\text{H}}_8{\text{ molecules/cm}}^3\end{array}$

Expression for ideal gas equation is as follows:

  $PV=nRT$

Where,

• $P$ is pressure of gas.
• $V$ is volume of gas.
• $n$ is amount or moles of gas.
• $R$ is universal gas constant.
• $T$ is absolute temperature of gas.

Rearrange above equation to calculate value of $P$ for volume of ${\text{C}}_7{\text{H}}_8$ .

  $V=\frac{nRT}{P}$

Value of $P$ is $\text{748 torr}$ .

Value of $n$ is $1.66\times {10}^{-9}\text{ mol}$ .

Value of $T$ is $296\text{K}$ .

Value of $R$ is $0.08206\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ .

Substitute the value in above equation.

  $\begin{array}{c}V=\frac{nRT}{P}\\ =\left(\frac{\left(1.66\times {10}^{-9}\text{ mol}\right)\left(0.08206\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(296\text{K}\right)}{\text{748 torr}}\right)\left(\frac{760\text{ torr}}{1\text{ atm}}\right)\\ =4.10\times {10}^{-8}\text{ L}\end{array}$

Expression of $\text{ppmv}$ for ${\text{C}}_7{\text{H}}_8$ is as follows:

  ${\text{ppmv of C}}_7{\text{H}}_8=\left(\frac{{\text{Volume of C}}_7{\text{H}}_8\text{ at STP}}{\text{Total volume of air at STP}}\right)\left({10}^6\right)$

Where,

• $X$ is trace compound.

Value of volume of ${\text{C}}_7{\text{H}}_8$ is $4.10\times {10}^{-8}\text{ L}$ .

Value of total volume equivalent of ${\text{C}}_7{\text{H}}_8$ is $3.00\text{ L}$ .

Substitute the value in above equation.

  $\begin{array}{c}{\text{ppmv of C}}_7{\text{H}}_8=\left(\frac{{\text{Volume of C}}_7{\text{H}}_8\text{ at STP}}{\text{Total volume of air at STP}}\right)\left({10}^6\right)\\ =\left(\frac{4.10\times {10}^{-8}\text{ L}}{3.00\text{ L}}\right)\left({10}^6\right)\\ =1.37\times {10}^{-2}\text{ ppmv}\end{array}$

Conclusion

Mixing ratio of benzene and toluene is $9.47\times {10}^{-3}\text{ ppmv}$ and $1.37\times {10}^{-2}\text{ ppmv}$ respectively. Number of molecule of benzene and toluene per cubic centimeter is $2.31\times {10}^{11}{\text{ molecules/cm}}^3$ and $3.21\times {10}^{11}{\text{ molecules/cm}}^3$ respectively.