Chapter 5, Problem 130E

Interpretation Introduction

Interpretation: For the given data, mixing ratio and number of molecules per cubic centimeter for both benzene and toluene should be determined.

Concept introduction:

• Mixing ratio is used to express the concentration of trace compounds in air. Mixing ratios are often expressed as ppmv (parts per million volume);

  $ppmvofX=\frac{volumeofXatSTP}{totalvolumeofairatSTP}\times 10^6$

• Number of moles of a substance,

  From its given mass is,

$\text{Number of moles}\text{=}\frac{\text{Given}\text{mass}}{\text{Molecular}\text{mass}}$

$\text{Number}\text{of}\text{molecules}\text{=Number}\text{of}\text{moles}\text{×}\frac{\text{6}{\text{.022×10}}^{\text{23}}\text{molecules}}{\text{mol}}$

• By combining the three gaseous laws namely Boyle’s law, Charles’s law and Avogadro’s law a combined gaseous equation is obtained. This combined gaseous equation is called Ideal gas law.

According to ideal gas law,

$PV=nRT$

By rearranging the equation, unknown volume can be determined as,

$V=\frac{nRT}{P}$

Where,

P = pressure in atmospheres

V= volumes in liters

n = number of moles

R =universal gas constant ( $\text{0}\text{.08206L×atm/K×mol}$)

T = temperature in kelvins

Answer to Problem 130E

Answer

• Mixing ratio of benzene  $=9.47\times 10^{-3}ppmv$
• Mixing ratio of toluene   $=1.37\times 10^{-2}ppmv$
• Number of molecules per cubic centimeter for benzene 

$=2.31\times 10^{11}moleculesbenzene/c{m}^3$

• Number of molecules per cubic centimeter for toluene

$=3.33\times 10^{11}moleculestoluene/c{m}^3$

Explanation of Solution

Explanation

• To determine: The mixing ratio of benzene

Mixing ratio of benzene  $=9.47\times 10^{-3}ppmv$

$\begin{array}{l}\text{For}\text{benzene,}\\ \text{To}\text{calculate}\text{themixing}\text{ratio,}\text{number}\text{of}\text{moles}\text{and}\text{volume}\text{of}\text{benzene}\text{should}\text{bedetermined}\text{.}\\ \text{Number of moles}\text{=}\frac{\text{Given}\text{mass}}{\text{Molecular}\text{mass}}\\ \text{Given}\text{mass}\text{=89}\text{.6}{\text{×10}}^{\text{-9}}\text{g}\\ \text{Molecular}\text{mass=78}\text{.11g}\\ \text{Number}\text{of}\text{moles}\left({\text{n}}_{\text{benzene}}\right)\text{=}\frac{\text{89}\text{.6}{\text{×10}}^{\text{-9}}\text{g}}{\text{78}\text{.11g}}\text{=}\text{1}\text{.15}{\text{×10}}^{\text{-9}}\text{mol}\text{benzene}\\ \text{Volume}\text{of}\text{benzene}\text{=}\frac{{\text{n}}_{\text{benzene}}\text{×}\text{R×}\text{T}}{\text{P}}\\ \text{R=0}\text{.08206L×atm/K×mol}\\ \text{T=}\text{23°C=296K}\text{Since,}\text{K}\text{=}\text{°C+273}\\ \text{=}\text{23°C+273}\\ \text{=296K}\\ \text{P=}\text{748}\text{torr=}\text{0}\text{.99atm}\text{Since,1}\text{atm=760torr}\\ \text{748torr=}\frac{\text{748}}{\text{760}}\text{atm}\\ \text{=}\text{0}\text{.99atm}\\ {\text{V}}_{\text{benzene}}\text{=}\frac{\text{1}\text{.15}{\text{×10}}^{\text{-9}}\text{mol×296K}\text{×}\frac{\text{0}\text{.08206 L}\text{atm}}{\text{K}\text{mol}}}{\text{0}\text{.99atm}}\\ \text{=}\text{2}\text{.84}\text{×}{\text{10}}^{\text{-8}}\text{L}\\ \text{Mixing}\text{ratio}\text{=}\frac{\text{volume}\text{of}\text{X}\text{at}\text{STP}}{\text{total}\text{volume}\text{of}\text{air}\text{at}\text{STP}}{\text{×10}}^{\text{6}}\\ {\text{V}}_{\text{benzene}}\text{=}\text{2}\text{.84}\text{×}{\text{10}}^{\text{-8}}\text{L}\\ \text{Total}\text{volume of}\text{the}\text{sample}\text{=}\text{3}\text{.00L}\\ \text{Mixing}\text{ratio}\text{=}\frac{\text{2}\text{.84}\text{×}{\text{10}}^{\text{-8}}\text{L}}{\text{3}\text{.00L}}\times {\text{10}}^6\text{=9}\text{.47}{\text{×10}}^{\text{-3}}\text{ppmv}\\ \\ \end{array}$

• To determine: Number of molecules per cubic centimeter of benzene.

Number of molecules per cubic centimeter for benzene

$=2.31\times 10^{11}moleculesbenzene/c{m}^3$

$\begin{array}{l}\\ \text{Number}\text{of}\text{molecules}\text{per}\text{cubic}\text{centimetre}\text{=Number}\text{of}\text{moles}\text{×}\frac{\text{6}{\text{.022×10}}^{\text{23}}\text{molecules}}{\text{mol}}\\ \text{Total}\text{volume}\text{of}\text{the}\text{sample=3}\text{.0L}\\ \text{Number}\text{of}\text{moles}\text{per}\text{volume}\text{of}\text{the}\text{sample=}\frac{\text{1}{\text{.15×10}}^{\text{-9}}\text{mol}\text{benzene}}{\text{3}\text{.00L}}\\ \text{Since,}\text{1L=1000}{\text{cm}}^{\text{3}}\\ \text{Number}\text{of}\text{moles}\text{per}\text{cubic}\text{centimetre}\text{=}\frac{\text{1}{\text{.15×10}}^{\text{-9}}\text{mol}\text{benzene}}{\text{3}\text{.00L}}\text{×}\frac{\text{1}}{\text{1000}{\text{cm}}^{\text{3}}}\\ \text{Number}\text{of}\text{molecules}\text{per}\text{cubic}\text{centimetre}\text{=}\frac{\text{1}{\text{.15×10}}^{\text{-9}}\text{mol}\text{benzene}}{\text{3}\text{.00L}}\text{×}\frac{\text{1}}{\text{1000}{\text{cm}}^{\text{3}}}\text{×}\frac{\text{6}{\text{.022×10}}^{\text{23}}\text{molecules}}{\text{mol}}\\ \text{=}\text{2}\text{.31}{\text{×10}}^{\text{11}}\text{molecules}{\text{benzene/cm}}^{\text{3}}\\ \end{array}$

• To determine: The mixing ratio of toluene

Mixing ratio of toluene $=1.37\times 10^{-2}ppmv$

$\begin{array}{l}\text{For}\text{toluene,}\\ \text{To}\text{calculate}\text{themixing}\text{ratio,}\text{number}\text{of}\text{moles}\text{and}\text{volume}\text{of}\text{benzene}\text{should}\text{bedetermined}\text{.}\\ \text{Number of moles}\text{=}\frac{\text{Given}\text{mass}}{\text{Molecular}\text{mass}}\\ \text{Given}\text{mass}\text{=153}{\text{×10}}^{\text{-9}}\text{g}\\ \text{Molecular}\text{mass=92}\text{.13g}\\ \text{Number}\text{of}\text{moles}\left({\text{n}}_{\text{toluene}}\right)\text{=}\frac{\text{153}{\text{×10}}^{\text{-9}}\text{g}}{\text{92}\text{.13g}}\text{=}\text{1}\text{.66}{\text{×10}}^{\text{-9}}\text{mol}\text{toluene}\\ \text{Volume}\text{of}\text{toluene}\text{=}\frac{{\text{n}}_{\text{toluene}}\text{×}\text{R×}\text{T}}{\text{P}}\\ \text{R=0}\text{.08206L×atm/K×mol}\\ \text{T=}\text{23°C=296K}\text{Since,}\text{K}\text{=}\text{°C+273}\\ \text{=}\text{23°C+273}\\ \text{=296K}\\ \text{P=}\text{748}\text{torr=}\text{0}\text{.99atm}\text{Since,1}\text{atm=760torr}\\ \text{748torr=}\frac{\text{748}}{\text{760}}\text{atm}\\ \text{=}\text{0}\text{.99atm}\\ {\text{V}}_{\text{toluene}}\text{=}\frac{\text{1}\text{.66}{\text{×10}}^{\text{-9}}\text{mol×296K}\text{×}\frac{\text{0}\text{.08206 L}\text{atm}}{\text{K}\text{mol}}}{\text{0}\text{.99atm}}\\ \text{=}\text{4}\text{.10}\text{×}{\text{10}}^{\text{-8}}\text{L}\\ \text{Mixing}\text{ratio}\text{=}\frac{\text{volume}\text{of}\text{X}\text{at}\text{STP}}{\text{total}\text{volume}\text{of}\text{air}\text{at}\text{STP}}{\text{×10}}^{\text{6}}\\ {\text{V}}_{\text{toluene}}\text{=}\text{4}\text{.10}\text{×}{\text{10}}^{\text{-8}}\text{L}\\ \text{Total}\text{volume of}\text{the}\text{sample}\text{=}\text{3}\text{.00L}\\ \text{Mixing}\text{ratio}\text{=}\frac{\text{4}\text{.10}\text{×}{\text{10}}^{\text{-8}}\text{L}}{\text{3}\text{.00L}}{\text{×10}}^{\text{6}}\text{=}\text{1}\text{.37}{\text{×10}}^{\text{-2}}\text{ppmv}\\ \\ \end{array}$

• To determine: Number of molecules per cubic centimeter of toluene.

Number of molecules per cubic centimeter for toluene

$=3.33\times 10^{11}moleculestoluene/c{m}^3$

$\begin{array}{l}\\ \text{Number}\text{of}\text{molecules}\text{per}\text{cubic}\text{centimetre}\text{=Number}\text{of}\text{moles}\text{×}\frac{\text{6}{\text{.022×10}}^{\text{23}}\text{molecules}}{\text{mol}}\\ \text{Total}\text{volume}\text{of}\text{the}\text{sample=3}\text{.0L}\\ \text{Number}\text{of}\text{moles}\text{per}\text{volume}\text{of}\text{the}\text{sample=}\frac{\text{1}{\text{.66×10}}^{\text{-9}}\text{mol}\text{toluene}}{\text{3}\text{.00L}}\\ \text{Since,}\text{1L=1000}{\text{cm}}^{\text{3}}\\ \text{Number}\text{of}\text{moles}\text{per}\text{cubic}\text{centimetre}\text{=}\frac{\text{1}{\text{.66×10}}^{\text{-9}}\text{mol}\text{toluene}}{\text{3}\text{.00L}}\text{×}\frac{\text{1}}{\text{1000}{\text{cm}}^{\text{3}}}\\ \text{Number}\text{of}\text{molecules}\text{per}\text{cubic}\text{centimetre}\text{=}\frac{\text{1}{\text{.66×10}}^{\text{-9}}\text{mol}\text{toluene}}{\text{3}\text{.00L}}\text{×}\frac{\text{1}}{\text{1000}{\text{cm}}^{\text{3}}}\text{×}\frac{\text{6}{\text{.022×10}}^{\text{23}}\text{molecules}}{\text{mol}}\\ \text{=}\text{3}\text{.33}{\text{×10}}^{\text{11}}\text{molecules}{\text{toluene/cm}}^{\text{3}}\\ \end{array}$

Conclusion

Conclusion

Mixing ratio and number of molecules per cubic centimeter for both benzene and toluene is determined on the basis of respective equations.