Methyl Alcohol (CH3OH) and ethyl alcohol (C2H5OH) form ideal solutions at 50°C, the vapor pressure of pure methyl alcohol is 0.529 atm, and the vapor pressure of pure ethyl alcohol is 0.292 atm. What is the vapor pressure at 50°C of a solution containing 24.00 g of methyl alcohol and 5.76 g of ethyl alcohol?

First, we need to find the mole fraction of each component in the solution.

Molar mass of methyl alcohol (CH3OH) = 12.01 g/mol (C) + 3 * 1.01 g/mol (H) + 1 * 16.00 g/mol (O) + 1 * 1.01 g/mol (H) = 32.04 g/mol

Moles of methyl alcohol = 24.00 g / 32.04 g/mol = 0.749 moles

Molar mass of ethyl alcohol (C2H5OH) = 2 * 12.01 g/mol (C) + 5 * 1.01 g/mol (H) + 1 * 16.00 g/mol (O) + 1 * 1.01 g/mol (H) = 46.07 g/mol

Moles of ethyl alcohol = 5.76 g / 46.07 g/mol = 0.125 moles

Total moles in the solution = moles of methyl alcohol + moles of ethyl alcohol = 0.749 + 0.125 = 0.874 moles

Mole fraction of methyl alcohol = moles of methyl alcohol / total moles = 0.749 / 0.874 = 0.857

Mole fraction of ethyl alcohol = moles of ethyl alcohol / total moles = 0.125 / 0.874 = 0.143

Now, we can use the mole fractions and the ideal solution assumption to calculate the vapor pressure of each component in the solution:

Methyl alcohol:
P(CH3OH) = mole fraction of CH3OH * vapor pressure of pure CH3OH = 0.857 * 0.529 atm = 0.453 atm

Ethyl alcohol:
P(C2H5OH) = mole fraction of C2H5OH * vapor pressure of pure C2H5OH = 0.143 * 0.292 atm = 0.042 atm

Finally, we can find the total vapor pressure of the solution by adding the vapor pressures of each component:

P(total) = P(CH3OH) + P(C2H5OH) = 0.453 atm + 0.042 atm = 0.495 atm

So the vapor pressure of the solution at 50°C is 0.495 atm.

To determine the vapor pressure of the solution, we need to calculate the mole fraction of each component and then use Raoult's Law.

Step 1: Calculate the moles of each component.
Mole of methyl alcohol (CH3OH):
Number of moles = mass / molar mass
Molar mass of CH3OH = 12.01 + (3 x 1.01) + 16.00 = 32.04 g/mol
Moles of CH3OH = 24.00 g / 32.04 g/mol

Mole of ethyl alcohol (C2H5OH):
Number of moles = mass / molar mass
Molar mass of C2H5OH = (2 x 12.01) + (5 x 1.01) + 16.00 = 46.07 g/mol
Moles of C2H5OH = 5.76 g / 46.07 g/mol

Step 2: Calculate the mole fraction of each component.
Mole fraction of CH3OH = Moles of CH3OH / (Moles of CH3OH + Moles of C2H5OH)
Mole fraction of CH3OH = (24.00 g / 32.04 g/mol) / [(24.00 g / 32.04 g/mol) + (5.76 g / 46.07 g/mol)]

Mole fraction of C2H5OH = Moles of C2H5OH / (Moles of CH3OH + Moles of C2H5OH)
Mole fraction of C2H5OH = (5.76 g / 46.07 g/mol) / [(24.00 g / 32.04 g/mol) + (5.76 g / 46.07 g/mol)]

Step 3: Use Raoult's Law to calculate the vapor pressure of the solution.
Raoult's Law states that the vapor pressure of a solution is the mole fraction of the component multiplied by the vapor pressure of the pure component.
Vapor pressure of solution = Mole fraction of CH3OH x Vapor pressure of pure CH3OH + Mole fraction of C2H5OH x Vapor pressure of pure C2H5OH

Vapor pressure of solution = (Mole fraction of CH3OH x 0.529 atm) + (Mole fraction of C2H5OH x 0.292 atm)

Now you can substitute the values we calculated into the equation to find the vapor pressure of the solution at 50°C.

To calculate the vapor pressure of the solution containing methyl alcohol and ethyl alcohol, we can use Raoult's Law. According to Raoult's Law, the vapor pressure of a component in an ideal solution is equal to the product of the mole fraction of that component and its vapor pressure in the pure state.

Step 1: Calculate the moles of methyl alcohol (CH3OH) and ethyl alcohol (C2H5OH) in the solution.
Molar mass of CH3OH = 32.04 g/mol
Molar mass of C2H5OH = 46.07 g/mol

Moles of CH3OH = Mass of CH3OH / Molar mass of CH3OH
= 24.00 g / 32.04 g/mol
= 0.749 mol

Moles of C2H5OH = Mass of C2H5OH / Molar mass of C2H5OH
= 5.76 g / 46.07 g/mol
= 0.125 mol

Step 2: Calculate the mole fraction of methyl alcohol (CH3OH) and ethyl alcohol (C2H5OH) in the solution.

Mole fraction of CH3OH = Moles of CH3OH / Total moles in the solution
= 0.749 mol / (0.749 mol + 0.125 mol)
= 0.857

Mole fraction of C2H5OH = Moles of C2H5OH / Total moles in the solution
= 0.125 mol / (0.749 mol + 0.125 mol)
= 0.143

Step 3: Calculate the vapor pressure of methyl alcohol (CH3OH) and ethyl alcohol (C2H5OH) in the solution using Raoult's Law.

Vapor pressure of CH3OH in the solution = Mole fraction of CH3OH * Vapor pressure of CH3OH in pure state
= 0.857 * 0.529 atm
= 0.454 atm

Vapor pressure of C2H5OH in the solution = Mole fraction of C2H5OH * Vapor pressure of C2H5OH in pure state
= 0.143 * 0.292 atm
= 0.042 atm

Step 4: Calculate the total vapor pressure of the solution.

Total vapor pressure = Vapor pressure of CH3OH in the solution + Vapor pressure of C2H5OH in the solution
= 0.454 atm + 0.042 atm
= 0.496 atm

Therefore, the vapor pressure at 50°C of the solution containing 24.00 g of methyl alcohol and 5.76 g of ethyl alcohol is 0.496 atm.