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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);
ppmv of X = volume of X at STPtotal volume of air at STP×106
- Number of moles of a substance,
From its given mass is,
Number of moles = Given massMolecular mass
Number of molecules =Number of moles ×6.022×1023moleculesmol
- 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 calledIdeal gas law .
According to ideal gas law,
PV=nRT
By rearranging the equation, unknown volume can be determined as,
V= n R TP
Where,
P = pressure in atmospheres
V= volumes in liters
n = number of moles
R =universal gas constant ( 0.08206L×atm/K×mol)
T = temperature in kelvins
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Answer to Problem 130E
Answer
- Mixing ratio of benzene = 9.47 × 10-3ppmv
- Mixing ratio of toluene = 1.37 × 10-2ppmv
- Number of molecules per cubic centimeter for benzene
= 2.31 ×1011molecules benzene/cm3
- Number of molecules per cubic centimeter for toluene
= 3.33 ×1011 molecules toluene/cm3
Explanation of Solution
Explanation
- To determine: The mixing ratio of benzene
Mixing ratio of benzene = 9.47 × 10-3ppmv
For benzene,To calculate themixing ratio, number of moles and volume of benzene should bedetermined.Number of moles = Given massMolecular massGiven mass =89.6 ×10-9 gMolecular mass=78.11gNumber of moles(nbenzene)= 89.6 ×10-9 g78.11g= 1.15 ×10-9mol benzeneVolume of benzene =nbenzene × R× TPR=0.08206L×atm/K×molT= 23°C=296K Since, K = °C+273 = 23°C+273 =296KP= 748 torr= 0.99atm Since,1 atm=760torr 748torr= 748 760 atm = 0.99atmVbenzene=1.15 ×10-9mol×296K ×0.08206 L atmK mol0.99atm = 2.84 × 10-8LMixing ratio = volume of X at STPtotal volume of air at STP×106Vbenzene= 2.84 × 10-8LTotal volume of the sample = 3.00LMixing ratio =2.84 × 10-8L3.00L ×106=9.47 ×10-3ppmv
- To determine: Number of molecules per cubic centimeter of benzene.
Number of molecules per cubic centimeter for benzene
= 2.31 ×1011molecules benzene/cm3
Number of molecules per cubic centimetre =Number of moles ×6.022×1023moleculesmolTotal volume of the sample=3.0L Number of moles per volume of the sample= 1.15×10-9 mol benzene3.00L Since, 1L=1000 cm3Number of moles per cubic centimetre = 1.15×10-9 mol benzene3.00L×11000 cm3Number of molecules per cubic centimetre =1.15×10-9 mol benzene3.00L×11000 cm3×6.022×1023moleculesmol = 2.31 ×1011molecules benzene/cm3
- To determine: The mixing ratio of toluene
Mixing ratio of toluene = 1.37 × 10-2ppmv
For toluene,To calculate themixing ratio, number of moles and volume of benzene should bedetermined.Number of moles = Given massMolecular massGiven mass =153 ×10-9 gMolecular mass=92.13gNumber of moles(ntoluene)= 153 ×10-9 g92.13g= 1.66 ×10-9mol tolueneVolume of toluene =ntoluene × R× TPR=0.08206L×atm/K×molT= 23°C=296K Since, K = °C+273 = 23°C+273 =296KP= 748 torr= 0.99atm Since,1 atm=760torr 748torr= 748 760 atm = 0.99atmVtoluene=1.66 ×10-9mol×296K ×0.08206 L atmK mol0.99atm = 4.10 × 10-8LMixing ratio = volume of X at STPtotal volume of air at STP×106Vtoluene= 4.10 × 10-8LTotal volume of the sample = 3.00LMixing ratio =4.10 × 10-8L3.00L×106 = 1.37 ×10-2ppmv
- To determine: Number of molecules per cubic centimeter of toluene.
Number of molecules per cubic centimeter for toluene
= 3.33 ×1011molecules toluene/cm3
Number of molecules per cubic centimetre =Number of moles ×6.022×1023moleculesmolTotal volume of the sample=3.0L Number of moles per volume of the sample= 1.66×10-9 mol toluene3.00L Since, 1L=1000 cm3Number of moles per cubic centimetre = 1.66×10-9 mol toluene3.00L×11000 cm3Number of molecules per cubic centimetre =1.66×10-9 mol toluene3.00L×11000 cm3×6.022×1023moleculesmol = 3.33 ×1011molecules toluene/cm3
Conclusion
Mixing ratio and number of molecules per cubic centimeter for both benzene and toluene is determined on the basis of respective equations.
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Chapter 5 Solutions
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