Question

Mixtures of benzene and cyclohexane exhibit ideal behavior. A solution was created containing 1.5 moles of liquid benzene and 2.5 moles of liquid cyclohexane, and at 50°C the total vapor pressure of the solution was measured to be 340 torr. Another solution was created containing 1.5 moles of liquid benzene and 3.5 moles of liquid cyclohexane, and at 50°C the measured total vapor pressure was 370 torr.

(A) Calculate the vapor pressures of pure benzene and pure cyclohexane at 50°C.

Answer 1

To find the vapor pressures of pure benzene and pure cyclohexane at 50°C, Raoult's Law is applied using a system of equations based on the given total vapor pressures for two different composition mixtures. Solving the system reveals the vapor pressures of the pure substances.

Step-by-step explanation:

To calculate the vapor pressures of pure benzene and pure cyclohexane at 50°C, we can use Raoult's Law which states that the partial vapor pressure of a component in an ideal mixture is equal to the mole fraction of the component in the liquid phase multiplied by its pure vapor pressure. The total vapor pressure of the mixture is then the sum of the partial vapor pressures of the components.

The total vapor pressure for the first solution is 340 torr, which consists of 1.5 moles of benzene and 2.5 moles of cyclohexane. For the second solution, the total vapor pressure is 370 torr with 1.5 moles of benzene and 3.5 moles of cyclohexane. Let Pb and Pc represent the vapor pressures of pure benzene and cyclohexane respectively. The mole fraction of each component in the solution is used to calculate the vapor pressure contributions according to Raoult's Law: Ptotal = (mole fraction of benzene x Pb) + (mole fraction of cyclohexane x Pc).

By setting up a system of equations using the vapor pressure data from both solutions and solving for Pb and Pc, we can obtain the vapor pressures of the pure components.

Answer 2

Vapour pressure of cyclohexane at 50°C is 490torr

Vapour pressure of benzene at 50°C is 90torr

Step-by-step explanation:

Using Raoult's law, pressure of a solution is defined by the sum of the product sbetween mole fraction of both solvents and pressure of pure solvents.

`P_(solution) = X_(A)P^0_(A)+X_(B)P^0_(B)`

In the first solution:

`X_(cyclohexane)=(2.5mol)/(2.5mol+1.5mol) =0.625`

`X_(benzene)=(1.5mol)/(2.5mol+1.5mol) =0.375`

`340torr = 0.625P^0_(A)+0.375P^0_(B)` (1)

For the second equation:

`X_(cyclohexane)=(3.5mol)/(3.5mol+1.5mol) =0.700`

`X_(benzene)=(1.5mol)/(3.5mol+1.5mol) =0.300`

`370torr = 0.700P^0_(A)+0.300P^0_(B)`(2)

Replacing (2) in (1):

`340torr = 0.625P^0_(A)+0.375(1233.3-2.333P^0_(A))`

`340torr = 0.625P^0_(A)+462.5-0.875P^0_(A)`

-122.5torr = -0.250P°A

`P^0_(A) = 490 torr`

Vapour pressure of cyclohexane at 50°C is 490torr

And for benzene:

`370torr = 0.700*490torr+0.300P^0_(B)`

`P^0_(B)=90torr`

Vapour pressure of benzene at 50°C is 90torr