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Answer :
Final answer:
To determine the vapor pressure of a methanol-ethanol mixture at 293 K, use the molecular weights of both alcohols to calculate their moles, find their mole fractions, and apply Raoult's law to find the partial pressures. The sum of the partial pressures gives the total vapor pressure of the mixture.
Explanation:
The student is asking about calculating the vapor pressure of a mixture at 293 K using Raoult's law. To calculate the total vapor pressure of the solution, we must first determine the mole fractions of both components. We do this by knowing the molecular weights of ethanol (C₂H₅OH) and methanol (CH₃OH), which are about 46.07 g/mol and 32.04 g/mol respectively. Then, using the amounts given (80 g of ethanol and 97 g of methanol), we determine the moles of each:
80 g ethanol ÷ 46.07 g/mol = 1.737 moles of ethanol
97 g methanol ÷ 32.04 g/mol = 3.028 moles of methanol
Now we calculate the total moles in the solution and mole fractions:
Total moles = 1.737 moles + 3.028 moles = 4.765 moles
Mole fraction of ethanol = 1.737 ÷ 4.765
Mole fraction of methanol = 3.028 ÷ 4.765
Next, we use Raoult's law to calculate the partial vapor pressure of each component in the solution:
Partial pressure of ethanol = Mole fraction of ethanol × Vapor pressure of pure ethanol
Partial pressure of methanol = Mole fraction of methanol × Vapor pressure of pure methanol
Finally, we add the partial pressures to get the total vapor pressure of the mixture:
Total vapor pressure = Partial pressure of ethanol + Partial pressure of methanol
Using the given vapor pressures of 97.7 Torr for methanol and 44.6 Torr for ethanol at 293 K, the calculation is conducted with the appropriate mole fractions.
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