Final answer:
Using Raoult's Law and the given vapor pressures, along with molar mass calculations, the vapor pressure of ethanol above the solution of methanol and ethanol mixed in equal masses at 20°C is approximately 18.08 torr.
Step-by-step explanation:
The vapor pressure of ethanol in a solution made by mixing equal masses of methanol and ethanol can be determined using Raoult's Law, which states that the vapor pressure of a component in an ideal mixture is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution. To find the vapor pressure of ethanol, you first need to calculate the mole fractions of methanol and ethanol in the solution.
To calculate the mole fraction, you need the moles of each substance. The molar mass of methanol (CH3OH) is approximately 32.04 g/mol, and the molar mass of ethanol (C2H5OH) is approximately 46.07 g/mol. Using these values, you can calculate the moles of each alcohol. Then, using the mole fractions and the given vapor pressures of methanol (94 torr) and ethanol (44 torr) at 20°C, you can apply Raoult's Law to find the vapor pressure of ethanol above the solution.
Here's an example calculation:
- Calculate moles of methanol: 50.0 g / 32.04 g/mol = 1.56 moles
- Calculate moles of ethanol: 50.0 g / 46.07 g/mol = 1.09 moles
- Calculate total moles in the solution: 1.56 moles + 1.09 moles = 2.65 moles
- Calculate mole fraction of ethanol: 1.09 moles / 2.65 moles = 0.411
- Apply Raoult's Law for ethanol vapor pressure: 0.411 * 44 torr = 18.08 torr
The vapor pressure of ethanol above the solution at 20°C would be approximately 18.08 torr.