A mixture of three miscible liquids (A, B and C) at 25oC has a total vapour pressure of
0.35 atm due to the components of the mixture. The mole fractions (X) of A and B are
0.25 and 0.50, respectively, and these two pure liquids have vapour pressures Po
A = 0.30
atm and Po
B = 0.40 atm at 25oC. What is the vapour pressure of pure component C (Po
C)
at 25oC
To find the vapor pressure of component C (PoC) at 25oC, we can use the total vapor pressure of the mixture and the mole fractions of the other two components.
The total vapor pressure of the mixture (Ptotal) is given as 0.35 atm.
The mole fraction of component A (XA) is 0.25, and its pure liquid vapor pressure (PoA) is 0.30 atm.
The mole fraction of component B (XB) is 0.50, and its pure liquid vapor pressure (PoB) is 0.40 atm.
The mole fraction of component C (XC) can be calculated by subtracting the sum of the mole fractions of A and B from 1:
XC = 1 - XA - XB
XC = 1 - 0.25 - 0.50
XC = 0.25
Now, we can use Raoult's law, which states that the vapor pressure of a component in a mixture is directly proportional to its mole fraction:
Ptotal = XA * PoA + XB * PoB + XC * PoC
Substituting the given values and solving for PoC:
0.35 = 0.25 * 0.30 + 0.50 * 0.40 + 0.25 * PoC
Simplifying the equation:
0.35 = 0.075 + 0.20 + 0.25 * PoC
Rearranging the equation:
0.35 - 0.075 - 0.20 = 0.25 * PoC
0.075 = 0.25 * PoC
Solving for PoC:
PoC = 0.075 / 0.25
PoC = 0.30 atm
Therefore, the vapor pressure of pure component C (PoC) at 25oC is 0.30 atm.