
Calculate the mole fraction of methanol in the vapor phase at $ {63.5^0}C $ for an ideal solution containing $ 25.0g $ of pure methanol and $ 75.0g $ of pure ethanol?
Some data necessary to solve this:
Methanol MM is $ 32.0gmo{l^{ - 1}} $
Vapor pressure is $ 717.2orr $
Ethanol MM is $ 46.1gmo{l^{ - 1}} $
Vapor pressure is $ 400.0torr $
Answer
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Hint: The number of moles of methanol, and ethanol will be calculated from their masses and molar masses. The mole fraction of methanol, and ethanol will be calculated from the moles of ethanol, methanol and total moles. The pressure of methanol and ethanol can be calculated from pressure given and mole fraction. The mole fraction of methanol at vapor pressure can be determined from the ratio of pressure of methanol to total pressure.
Complete answer:
The amount of methanol is $ 25g $ and molar mass of methanol is $ 32gmo{l^{ - 1}} $
Moles of methanol will be $ {n_{MeOH}} = \dfrac{{25}}{{32}} = 0.78125moles $
The amount of ethanol is $ 75g $ and molar mass of methanol is $ 46gmo{l^{ - 1}} $
Moles of methanol will be $ {n_{EtOH}} = \dfrac{{75}}{{46}} = 1.6269moles $
Total moles of methanol and ethanol will be $ 0.78125 + 1.6269 = 2.40815moles $
Mole fraction of Methanol will be $ {\chi _{MeOH}} = \dfrac{{0.78125}}{{2.40815}} = 0.3244 $
Mole fraction of ethanol will be $ {\chi _{EtOH}} = \dfrac{{1.6269}}{{2.40815}} = 0.6756 $
According to Rouault’s law,
$ {P_{solution}} = \chi .{P^0} $
The pressure of methanol will be $ {P_{MeOH}} = 0.3244 \times 717.2torr = 232.66torr $
The pressure of ethanol will be $ {P_{EtOH}} = 0.6756 \times 400torr = 270.24torr $
The total pressure will be $ 232.66 + 270.24 = 502.9torr $
The mole fraction of methanol in vapour phase can be calculated from the ratio of the pressure of methanol total pressure.
$ {\chi _{MeO{H_{Vapor}}}} = \dfrac{{232.66}}{{502.9}} = 0.463 $
Thus, the mole fraction of methanol in vapor phase is $ 0.463 $ .
Note:
The mole fraction is the ratio of moles of solute to the total moles of solute and solvent. It has no units. The mole fraction of solute is different from the mole fraction of solute in the vapor pressure. Mole fraction in the vapor phase means the ratio of pressure of methanol and total pressure.
Complete answer:
The amount of methanol is $ 25g $ and molar mass of methanol is $ 32gmo{l^{ - 1}} $
Moles of methanol will be $ {n_{MeOH}} = \dfrac{{25}}{{32}} = 0.78125moles $
The amount of ethanol is $ 75g $ and molar mass of methanol is $ 46gmo{l^{ - 1}} $
Moles of methanol will be $ {n_{EtOH}} = \dfrac{{75}}{{46}} = 1.6269moles $
Total moles of methanol and ethanol will be $ 0.78125 + 1.6269 = 2.40815moles $
Mole fraction of Methanol will be $ {\chi _{MeOH}} = \dfrac{{0.78125}}{{2.40815}} = 0.3244 $
Mole fraction of ethanol will be $ {\chi _{EtOH}} = \dfrac{{1.6269}}{{2.40815}} = 0.6756 $
According to Rouault’s law,
$ {P_{solution}} = \chi .{P^0} $
The pressure of methanol will be $ {P_{MeOH}} = 0.3244 \times 717.2torr = 232.66torr $
The pressure of ethanol will be $ {P_{EtOH}} = 0.6756 \times 400torr = 270.24torr $
The total pressure will be $ 232.66 + 270.24 = 502.9torr $
The mole fraction of methanol in vapour phase can be calculated from the ratio of the pressure of methanol total pressure.
$ {\chi _{MeO{H_{Vapor}}}} = \dfrac{{232.66}}{{502.9}} = 0.463 $
Thus, the mole fraction of methanol in vapor phase is $ 0.463 $ .
Note:
The mole fraction is the ratio of moles of solute to the total moles of solute and solvent. It has no units. The mole fraction of solute is different from the mole fraction of solute in the vapor pressure. Mole fraction in the vapor phase means the ratio of pressure of methanol and total pressure.
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