Question

You have a solution of two volatile liquids, A and B (assume ideal behavior). Pure liquid A has a vapor pressure of 385.0 torr and pure liquid B has a vapor pressure of 104.0 torr at the temperature of the solution. The vapor at equilibrium above the solution has double the mole fraction of substance A as the solution does. What is the mole fraction of liquid A in the solution?

Answer 1

the mole fraction of A in solution is 0.32

Step-by-step explanation:

Let's consider a solution of two volatile liquids A and B

Vapor pressure of A is 385.0 torr

Vapor pressure of B is 104.0 torr

According to Dalton's Law

`P_T = P_A + P_B` ---------- Equation (1)

where
`P_A \ and \ P_B` are partial pressure of A and B respectively.

Using Roult's Law

`P_T = X_AP^0_A + X_BP^0_B` ---------- Equation (2)

where;

`P_T =` the total vapor pressure of the solution

`X_A \ and \ X_B` = mole fraction of A and B respectively

`P_A^0 \ and \ P_B^0` = vapor pressures of pure species of A and B

So;

`P_A = X_AP^0`

`P_B =X_BP^0_B`

Replacing our given values for
`P_A^0` and
`P_B^0` into equation (2) ; we have:

`P_T = X_A * 385 + X_B * 104`

The sum of mole fractions is equal to 1

`X_A + X_B = 1`

`X_B = 1 - X_A`

Hence;

`P_T = X_A *385 + X_B * 104`

`= 385 X_A +(1 -X_A) 104\\ \\= 385 X_A + 104 - 104 X_A`

`= 281 X_A + 104` ------ Equation (3)

It is given that mole fraction of vapor is equal to twice the mole fraction of liquid .

`(X^0_A)_(vapor) = 2(X^0_A)_(liquid)`

It is known that; a partial pressure of a component in gaseous phase is equal to mole fraction times the total pressures. As such;

`P_A = (X^v_A)_(vapor) * P_T`

`(X^v_A)_(vapor) = (P_A)/(P_T)`

`(P_A)/(P_T)= 2(X^L_A)_(vapor)` ------- Equation (4)

Partial Pressure is given as follows :

`P_A =X^L_A P^0_A`

Thus; from equation 4

`(X_A^LP_A^0)/(P_T)= 2 (X^L_A)_(liquid)`

`(P^0_A)/(P_T )= 2`

`(P_A^0)/(2)= P_T`

Replacing the above value into equation (8)

`(P^0_A)/(2)= 281 X_A + 104`

`(385)/(2)= 281 X_A +104`

192.5 = 281
`X_A + 104`

192.5 - 104 =
`281 \ X_A`

88.5 =
`281 \ X_A`

`X_A = (88.5)/(281)`

`X_A` = 0.32

Thus, the mole fraction of A in solution is 0.32