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22/09/2021
##chi_”benzene” = 0.34##
##chi_”toluene” = 0.66##
The idea here is that the vapor pressures of benzene and toluene will contribute to the total vapor pressure of the solution proportionally to their respective mole fraction – this is known as Raoult’s Law.
Mathematically, you can express this by the following equation
##P_”sol” = chi_”benzene” * P_”benzene”^@ + chi_”toluene” * P_”toluene”^@##, where
##P_”sol”## – the vapor pressure of the solution
##chi_”benzene”## – fraction of benzene
##P_”benzene”^@## – the vapor pressure of pure benzene
Now, mole fraction is defined as the ratio between the number of moles of a component of a solution and the total number of moles present in the solution.
Since you only have two components to this solution, benzene and toluene, you can say that
##chi_”benzene” + chi_”toluene” = 1##
Let’s say that the mole fraction of benzene is ##x## and that of toluene is ##y##. You can say that
##x = 1 -y##
and
##P_”total” = x * P_”benzne”^@ + (1-x) * P_”toluene”^@##
##40. color(red)(cancel(color(black)(“torr”))) = x * 75color(red)(cancel(color(black)(“torr”))) + (1-x) * 22color(red)(cancel(color(black)(“torr”)))##
##40. = 75x + 22 – 22x##
##40 – 22 = 53x implies x = 18/53 = 0.3396##
This means that you have
y = 1 – 0.3396 = 0.6604
The mole fractions of benzene and toluene in the solution will thus be – rounded to two
##chi_”benzene” = color(green)(0.34)” “## and ##” “chi_”toluene” = color(green)(0.66)##
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I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
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