The vapor pressure of pure ethanol at 20 degrees C is 44 mmHg, and the vapor pressure of pure methanol at the same temperature is 94 mmHg. In a solution of the two, the mole fraction of ethanol is .439; the mole fraction for methanol is .561. Calculate the vapor pressure of each component of the solution.

Pethanol = Xethanol*Poethanol

The methanol is done the same way.

Pethanol means partial pressure of the ethanol above the solution.

To calculate the vapor pressure of each component in the solution, we can use Raoult's law, which states that the vapor pressure of a component in a solution is proportional to its mole fraction.

Let's start by calculating the vapor pressure of ethanol in the solution using Raoult's law.

Vapor pressure of ethanol in the solution = mole fraction of ethanol * vapor pressure of pure ethanol

Vapor pressure of ethanol in the solution = 0.439 * 44 mmHg

Vapor pressure of ethanol in the solution = 19.316 mmHg

Now let's calculate the vapor pressure of methanol in the solution using Raoult's law.

Vapor pressure of methanol in the solution = mole fraction of methanol * vapor pressure of pure methanol

Vapor pressure of methanol in the solution = 0.561 * 94 mmHg

Vapor pressure of methanol in the solution = 52.834 mmHg

Therefore, the vapor pressure of ethanol in the solution is approximately 19.316 mmHg, and the vapor pressure of methanol in the solution is approximately 52.834 mmHg.

To calculate the vapor pressure of each component in the solution, we can use Raoult's Law, which states that the vapor pressure of a component in an ideal solution is proportional to its mole fraction.

First, calculate the vapor pressure of ethanol:
Vapor pressure of ethanol = Mole fraction of ethanol x Vapor pressure of pure ethanol

Substituting the given values into the equation:
Vapor pressure of ethanol = 0.439 x 44 mmHg
Vapor pressure of ethanol = 19.316 mmHg

Now, calculate the vapor pressure of methanol:
Vapor pressure of methanol = Mole fraction of methanol x Vapor pressure of pure methanol

Substituting the given values into the equation:
Vapor pressure of methanol = 0.561 x 94 mmHg
Vapor pressure of methanol = 52.834 mmHg

Therefore, in the solution, the vapor pressure of ethanol is 19.316 mmHg and the vapor pressure of methanol is 52.834 mmHg.