Answer:
the mole fraction of A in solution is 0.32
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
[tex]P_T = P_A + P_B[/tex] ---------- Equation (1)
where [tex]P_A \ and \ P_B[/tex] are partial pressure of A and B respectively.
Using Roult's Law
[tex]P_T = X_AP^0_A + X_BP^0_B[/tex] ---------- Equation (2)
where;
[tex]P_T =[/tex] the total vapor pressure of the solution
[tex]X_A \ and \ X_B[/tex] = mole fraction of A and B respectively
[tex]P_A^0 \ and \ P_B^0[/tex] = vapor pressures of pure species of A and B
So;
[tex]P_A = X_AP^0[/tex]
[tex]P_B =X_BP^0_B[/tex]
Replacing our given values for [tex]P_A^0[/tex] and [tex]P_B^0[/tex] into equation (2) ; we have:
[tex]P_T = X_A * 385 + X_B * 104[/tex]
The sum of mole fractions is equal to 1
[tex]X_A + X_B = 1[/tex]
[tex]X_B = 1 - X_A[/tex]
Hence;
[tex]P_T = X_A *385 + X_B * 104[/tex]
[tex]= 385 X_A +(1 -X_A) 104\\ \\= 385 X_A + 104 - 104 X_A[/tex]
[tex]= 281 X_A + 104[/tex] ------ Equation (3)
It is given that mole fraction of vapor is equal to twice the mole fraction of liquid .
[tex](X^0_A)_{vapor} = 2(X^0_A)_{liquid}[/tex]
It is known that; a partial pressure of a component in gaseous phase is equal to mole fraction times the total pressures. As such;
[tex]P_A = (X^v_A)_{vapor} * P_T[/tex]
[tex](X^v_A)_{vapor} = \frac{P_A}{P_T}[/tex]
[tex]\frac{P_A}{P_T}= 2(X^L_A)_{vapor}[/tex] ------- Equation (4)
Partial Pressure is given as follows :
[tex]P_A =X^L_A P^0_A[/tex]
Thus; from equation 4
[tex]\frac{X_A^LP_A^0}{P_T}= 2 (X^L_A)_{liquid}[/tex]
[tex]\frac{P^0_A}{P_T }= 2[/tex]
[tex]\frac{P_A^0}{2}= P_T[/tex]
Replacing the above value into equation (8)
[tex]\frac{P^0_A}{2}= 281 X_A + 104[/tex]
[tex]\frac{385}{2}= 281 X_A +104[/tex]
192.5 = 281[tex]X_A + 104[/tex]
192.5 - 104 = [tex]281 \ X_A[/tex]
88.5 = [tex]281 \ X_A[/tex]
[tex]X_A = \frac{88.5}{281}[/tex]
[tex]X_A[/tex] = 0.32
Thus, the mole fraction of A in solution is 0.32
