the pressure vapor of toluene is 40 torr at 31,8C and water is 35,3 torr. They are imiscibles. What is the vapor composition of a mixture toluene water, when x(TOLUENE)=0,25 ?
To determine the vapor composition of a mixture of toluene and water, you can use Raoult's Law. Raoult's Law states that the vapor pressure of an ideal mixture of two volatile components is equal to the vapor pressure of each component multiplied by its mole fraction.
Let's break down the given information:
- The vapor pressure of toluene (Pt) is 40 torr at 31.8°C.
- The vapor pressure of water (Pw) is 35.3 torr.
- The mixture is immiscible, meaning that the two liquids do not mix together to form a homogeneous solution.
- The mole fraction of toluene (x(TOLUENE)) is 0.25.
To find the vapor composition of the mixture, we need to calculate the mole fraction of water (x(WATER)) in the mixture. This can be done using the formula:
x(WATER) = 1 - x(TOLUENE)
Substituting the given value of x(TOLUENE) = 0.25:
x(WATER) = 1 - 0.25
= 0.75
Now, using Raoult's Law, we can calculate the vapor pressure of each component in the mixture:
P(TOLUENE) = Pt * x(TOLUENE)
= 40 torr * 0.25
= 10 torr
P(WATER) = Pw * x(WATER)
= 35.3 torr * 0.75
= 26.475 torr
Therefore, the vapor composition of the mixture of toluene and water is:
- Toluene vapor pressure: 10 torr
- Water vapor pressure: 26.475 torr