Final answer:
The partial pressures of ethanol and 1-propanol at 35°C are calculated using Raoult's Law. For ethanol with a mole fraction of 0.253, the partial pressure is approximately 34.23 mm Hg. For 1-propanol, with the remaining mole fraction, the partial pressure is approximately 28.09 mm Hg.
Step-by-step explanation:
To calculate the partial pressures of ethanol and 1-propanol at 35°C over a solution in which the mole fraction of ethanol is 0.253, we use Raoult's Law. Raoult's Law states that the partial vapor pressure of a component in a solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution. Given that the vapor pressure of pure ethanol at 35°C is not provided in the question, we will assume that it remains close to the provided value at 40°C, which is 0.178 atm. Similarly, we will use the 37.6 mmHg value provided for 1-propanol without any adjustments for temperature changes.
The partial pressure of ethanol (P₂₅₈OH) can be calculated as follows:
P₂₅₈OH = (mole fraction of C₂H₅OH) × (vapor pressure of pure C₂H₅OH)
= 0.253 × 0.178 atm
= 0.045034 atm
To convert atm to mmHg, we multiply by the conversion factor of 760 mmHg/atm:
P₂₅₈OH in mmHg = 0.045034 atm × 760 mmHg/atm
= 34.22584 mm Hg
The partial pressure of 1-propanol (Pʜ₇OH) can be calculated using the mole fraction of 1-propanol, which is 1 - 0.253 = 0.747, and the vapor pressure of 1-propanol:
Pʜ₇OH = (mole fraction of C₃H₇OH) × (vapor pressure of pure C₃H₇OH)
= 0.747 × 37.6 mmHg
= 28.0872 mm Hg
The partial pressures have been rounded to reflect the correct number of significant digits based on the initial data provided.