Oentane is a liquid with a vapor pressure of 512 Torr at 25C

[physicsss]

Oentane is a liquid with a vapor pressure of 512 Torr at 25C; at the same temperature, the vapor pressure of hexane is only 151 Torr. what composition must the liquid phase have if the gas hase composition is to have equal amounts of pentane and hexane?

Do I use p = i M R T to solve for the M, molar masses? How do I need to get started

 

[GCT]

No, you'll need to use the Rault's law

$$P=P_{pure~hexane}C_{hexane}+P_{pure~oentane}C_{oentane}$$

where C=molar composition of respective compound

 

[ravenprp]

To solve this problem, you will need to use the equation for Raoult's Law, which states that the vapor pressure of a solution is equal to the vapor pressure of the pure solvent multiplied by its mole fraction in the solution. In this case, the pure solvent is either pentane or hexane, and the solution is a mixture of the two.

To start, you will need to set up two equations using Raoult's Law. The first equation will represent the vapor pressure of the solution, which is equal to the sum of the vapor pressures of pentane and hexane:

P_solution = P_pentane + P_hexane

The second equation will represent the mole fractions of pentane and hexane in the solution, which must be equal in order to have equal amounts of each in the gas phase:

X_pentane = X_hexane

Now, you can substitute the given values for the vapor pressures of pentane and hexane (512 Torr and 151 Torr, respectively) and solve for the mole fractions:

P_solution = X_pentane * 512 Torr + X_hexane * 151 Torr

X_pentane = X_hexane

Next, you will need to use the ideal gas law to calculate the mole fractions of pentane and hexane in the gas phase. This can be done by using the given temperature (25C) and the known vapor pressures of pentane and hexane:

P_gas = X_pentane * P_pentane + X_hexane * P_hexane

P_gas = X_pentane * 512 Torr + X_hexane * 151 Torr

Finally, you can set the two equations for the mole fractions equal to each other and solve for the unknown mole fractions:

X_pentane * 512 Torr + X_hexane * 151 Torr = X_pentane * P_pentane + X_hexane * P_hexane

X_pentane = X_hexane

Solving this system of equations will give you the mole fractions of pentane and hexane in the liquid phase, which is the composition that will result in equal amounts of pentane and hexane in the gas phase. It is not necessary to use the molar masses in this problem.