
Benzene and toluene form an ideal solution over the entire range of composition. The vapor pressure of pure benzene and toluene at \[300K\] are \[50.71\,mm\] \[Hg\] and \[32.06\,mm\] \[Hg\] respectively. Calculate the mole fraction of benzene in vapor phase if \[80\,g\] of benzene is mixed with \[100\,g\] of toluene.
Answer
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Hint:An ideal solution is a blend wherein the molecules of various species are recognizable, however, in contrast to the ideal gas, the molecules in ideal solution apply force on each other. Raoult's law expresses that the vapor pressure of a solvent over a solution is equivalent to the vapor pressure of the pure solvent at a similar temperature scaled by the mole fraction of the solvent present.
Complete answer:
Step 1: To find out the molar mass of Benzene and Toluene
Molar Mass of Benzene \[\left( {{C}_{6}}{{H}_{6}} \right)=6\times 12+6\times 1=78\,g/mol\]
Molar Mass of Toluene \[\left( {{C}_{7}}{{H}_{8}} \right)=7\times 12+8\times 1=92\,g/mol\]
Step 2: To find out the Number of moles present in \[80\,g\] of Benzene and \[100\,g\] of Toluene
Number of moles in \[80g\]of Benzene \[=\dfrac{80}{78}=1.026mol\]
Number of moles in \[100g\]of Toluene\[=\dfrac{100}{92}=1.087mol\]
Step 3: Mole Fraction of Toluene and Benzene
Mole fraction of benzene, \[{{X}_{b}}=\dfrac{1.026}{1.026+1.087}=0.486\]
And Mole fraction of toluene, \[{{X}_{t}}=1-0.486=0.514\]
We have provided that
Vapor pressure of pure benzene \[{{P}_{b}}^{o}=50.71\,mm\,Hg\]
And Vapor pressure of pure toluene, \[{{P}_{t}}^{o}=32.06\,mm\,Hg\]
Now, partial vapor pressure of benzene, \[{{P}_{b}}={{P}_{b}}\times {{X}_{t}}\]
\[
=50.71\times 0.486 \\
=24.65\,mm\,Hg
\]
Now, partial vapor pressure of toluene, \[{{P}_{t}}={{P}_{t}}\times {{X}_{t}}\]
\[
{{P}_{t}}={{P}_{t}}^{o}\times {{X}_{t}}=32.06\times 0.514 \\
=16.48
\]
Hence, total vapor pressure \[=24.65+16.48=41.13\,mm\,Hg\]
Mole fraction of benzene in vapor phase \[=\dfrac{24.65}{42.13}=0.60\]
Therefore, the answer is \[0.60\]
Note:
It is essential to indicate the temperature while stating a vapor pressure since vapor pressures increment with temperature. Also, know that there are a few different units for pressure. The most widely recognized unit for vapor pressure is the \[torr\]. \[1torr=1\,mm\,Hg\] (one millimeter of mercury). Most materials have low vapor pressures.
Complete answer:
Step 1: To find out the molar mass of Benzene and Toluene
Molar Mass of Benzene \[\left( {{C}_{6}}{{H}_{6}} \right)=6\times 12+6\times 1=78\,g/mol\]
Molar Mass of Toluene \[\left( {{C}_{7}}{{H}_{8}} \right)=7\times 12+8\times 1=92\,g/mol\]
Step 2: To find out the Number of moles present in \[80\,g\] of Benzene and \[100\,g\] of Toluene
Number of moles in \[80g\]of Benzene \[=\dfrac{80}{78}=1.026mol\]
Number of moles in \[100g\]of Toluene\[=\dfrac{100}{92}=1.087mol\]
Step 3: Mole Fraction of Toluene and Benzene
Mole fraction of benzene, \[{{X}_{b}}=\dfrac{1.026}{1.026+1.087}=0.486\]
And Mole fraction of toluene, \[{{X}_{t}}=1-0.486=0.514\]
We have provided that
Vapor pressure of pure benzene \[{{P}_{b}}^{o}=50.71\,mm\,Hg\]
And Vapor pressure of pure toluene, \[{{P}_{t}}^{o}=32.06\,mm\,Hg\]
Now, partial vapor pressure of benzene, \[{{P}_{b}}={{P}_{b}}\times {{X}_{t}}\]
\[
=50.71\times 0.486 \\
=24.65\,mm\,Hg
\]
Now, partial vapor pressure of toluene, \[{{P}_{t}}={{P}_{t}}\times {{X}_{t}}\]
\[
{{P}_{t}}={{P}_{t}}^{o}\times {{X}_{t}}=32.06\times 0.514 \\
=16.48
\]
Hence, total vapor pressure \[=24.65+16.48=41.13\,mm\,Hg\]
Mole fraction of benzene in vapor phase \[=\dfrac{24.65}{42.13}=0.60\]
Therefore, the answer is \[0.60\]
Note:
It is essential to indicate the temperature while stating a vapor pressure since vapor pressures increment with temperature. Also, know that there are a few different units for pressure. The most widely recognized unit for vapor pressure is the \[torr\]. \[1torr=1\,mm\,Hg\] (one millimeter of mercury). Most materials have low vapor pressures.
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