Question

The vapor pressures of ethanol (C₂H₅OH) and 1-propanol (C₃H₇OH) at 35°C are 100 mmHg and 37.6 mmHg, respectively. Assuming ideal behavior, calculate the partial pressures of ethanol and 1-propanol at 35°C over a solution of ethanol in 1-propanol, in which the mole fraction of ethanol is 0.349.

Answer 1

To calculate the partial pressures of ethanol and 1-propanol in the solution, we can use Raoult's Law. First, calculate the mole fraction of ethanol. Then, use Raoult's Law to calculate the partial pressures of ethanol and 1-propanol.

Step-by-step explanation:

To calculate the partial pressures of ethanol and 1-propanol in the solution, we can use Raoult's Law. Raoult's Law states that the vapor pressure of a component in a solution is equal to the mole fraction of the component multiplied by its vapor pressure in the pure state.

First, calculate the mole fraction of ethanol using the given information:

Mole fraction of ethanol = moles of ethanol / (moles of ethanol + moles of 1-propanol)

Next, use Raoult's Law to calculate the partial pressures of ethanol and 1-propanol:

Partial pressure of ethanol = mole fraction of ethanol * vapor pressure of ethanol

Partial pressure of 1-propanol = mole fraction of 1-propanol * vapor pressure of 1-propanol