Answer :
Final answer:
The vapor pressure of a solution prepared by dissolving 5.00 g of benzoic acid in 100.00 g of ethyl alcohol at 35 ℃ is approximately 98.70 mmHg. This is calculated using Raoult's law, considering benzoic acid is nonvolatile, and only the mole fraction of ethyl alcohol affects the vapor pressure.
Explanation:
To calculate the vapor pressure of a solution prepared by dissolving benzoic acid in ethyl alcohol at 35 ℃, we can use Raoult's law. Raoult's law states that the vapor pressure of a solution is directly proportional to the mole fraction of the solvent. Since benzoic acid is nonvolatile and does not contribute to the vapor pressure, the vapor pressure of the solution will be that of ethyl alcohol multiplied by its mole fraction.
First, calculate the moles of ethyl alcohol and benzoic acid. The molar mass of ethyl alcohol (C2H6O) is about 46.07 g/mol, and the molar mass of benzoic acid (C7H6O2) is about 122.12 g/mol. For the 100.00 g of ethyl alcohol:
Moles of ethyl alcohol = 100.00 g / 46.07 g/mol = 2.171 moles
For the 5.00 g of benzoic acid:
Moles of benzoic acid = 5.00 g / 122.12 g/mol = 0.041 moles
The total moles in the solution is the sum of moles of ethyl alcohol and benzoic acid:
Total moles = 2.171 moles + 0.041 moles = 2.212 moles
The mole fraction of ethyl alcohol is then:
Mole fraction of ethyl alcohol = Moles of ethyl alcohol / Total moles = 2.171 moles / 2.212 moles = 0.9815
Now, we apply Raoult's law:
Vapor pressure of the solution = Mole fraction of ethyl alcohol × Vapor pressure of pure ethyl alcohol
Vapor pressure of the solution = 0.9815 × 100.5 mmHg = 98.70 mmHg
Therefore, the vapor pressure of the solution at 35 ℃ is approximately 98.70 mmHg.
