Question

methanol and ethanol have vapor pressures of 126.76 mmhg and 93.01 mmhg at 25°c, respectively. if you made a solution by mixing 50 g of pure methanol (CH₃OH) with 50 g of pure ethanol (CH₃CH₂OH), what would be the vapor pressure (in mmhg) of the ethanol above the solution.

Answer 1

The vapor pressure of ethanol in the mixture with methanol is calculated to be approximately 38.13 mmHg using Raoult's law after determining the mole fractions of each substance.

Step-by-step explanation:

To calculate the vapor pressure of ethanol in a mixture with methanol using Raoult's law, we must first find the mole fractions of each component. Given that the vapor pressure of pure ethanol at 25°C is 93.01 mmHg and that of methanol is 126.76 mmHg, we can determine the mole fraction of ethanol after mixing 50g of ethanol with 50g of methanol.

The molar mass of methanol (CH₃OH) is 32.04 g/mol and the molar mass of ethanol (C₂H₅OH) is 46.07 g/mol. Using these molar masses, we can find the number of moles of each:

• Moles of methanol = 50g / 32.04 g/mol = 1.560 moles
• Moles of ethanol = 50g / 46.07 g/mol = 1.085 moles

The total number of moles in the solution is 1.560 moles + 1.085 moles = 2.645 moles.

Now, calculate the mole fraction of ethanol:

Mole fraction of ethanol = Moles of ethanol / Total moles = 1.085 / 2.645 = 0.410

Finally, apply Raoult's law to determine the partial vapor pressure of ethanol above the solution:

Vapor pressure of ethanol = Mole fraction of ethanol × Vapor pressure of pure ethanol

Vapor pressure of ethanol = 0.410 × 93.01 mmHg = 38.13 mmHg

Therefore, the vapor pressure of ethanol above the solution is approximately 38.13 mmHg.