calculate the water vapor pressure of benzene at 0.80, 0.60, 0,40, and 0.20 mole fraction and toluene at 0.20, 0.40, 0.60, and 0.80 mole fraction.
You need to clarify the question There is no water vapor in benzene or toluene.
i don't know how to clarify the question. That's how it is on my paper.
To calculate the water vapor pressure of benzene and toluene at different mole fractions, we need to use Raoult's Law. Raoult's Law states that the vapor pressure of a component in an ideal solution is directly proportional to its mole fraction in the solution.
The formula for Raoult's Law is:
P = X * P₀
where P is the vapor pressure of the component, X is the mole fraction of the component, and P₀ is the vapor pressure of the pure component.
First, we need to determine the vapor pressure of water and benzene (P₀) and water and toluene (P₀) at the given temperature. Let's assume the temperature is constant.
Next, we can substitute the values of X and P₀ into the formula to calculate the water vapor pressure at each mole fraction.
For example, to calculate the water vapor pressure of benzene at 0.80 mole fraction:
1. Determine the vapor pressure of water (P₀) and benzene (P₀) at the given temperature.
2. Substitute X = 0.80 and P₀ into the formula: P = 0.80 * P₀
3. Calculate and obtain the value of P, which represents the water vapor pressure at a mole fraction of 0.80 benzene.
Repeat this process for the other mole fractions of benzene and toluene to obtain their respective water vapor pressures.