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?

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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.