Final answer:
The vapor pressure of the solution at 20°C containing 75.0 g of methanol and 25.0 g of ethanol is calculated using Raoult's law. After determining the moles and mole fractions for both components, their partial vapor pressures are found and summed up to obtain the total vapor pressure of 83.2 torr.
Step-by-step explanation:
To calculate the vapor pressure of a solution containing both methanol and ethanol, we need to use Raoult's law, which states that the partial vapor pressure of each component in an ideal mixture is equal to the mole fraction of the component multiplied by the vapor pressure of the pure component.
First, calculate the moles of methanol (CH3OH) and ethanol (CH3CH2OH) using their molecular weights (32.04 g/mol for methanol and 46.07 g/mol for ethanol).
- Moles of methanol = 75.0 g / 32.04 g/mol = 2.34 mol
- Moles of ethanol = 25.0 g / 46.07 g/mol = 0.54 mol
Next, find the total moles in the solution:
Total moles = 2.34 moles (methanol) + 0.54 moles (ethanol) = 2.88 moles
Now calculate the mole fractions:
- Mole fraction of methanol (Xmethanol) = 2.34 moles / 2.88 moles = 0.813
- Mole fraction of ethanol (Xethanol) = 0.54 moles / 2.88 moles = 0.187
Using Raoult's law, we can figure out the individual vapor pressures:
- Vapor pressure of methanol in the solution (Pmethanol) = Xmethanol * P°(CH3OH) = 0.813 * 92 torr = 74.8 torr
- Vapor pressure of ethanol in the solution (Pethanol) = Xethanol * P°(CH3CH2OH) = 0.187 * 45 torr = 8.4 torr
The total vapor pressure of the solution is the sum of these individual pressures:
Total vapor pressure = Pmethanol + Pethanol = 74.8 torr + 8.4 torr = 83.2 torr