Final answer:
The mole fraction of ethanol in the vapor phase from a mixture at 70°C and 2 atm depends on the vapor pressure of pure ethanol at that temperature, which must be known to apply Raoult's law alongside the given 70% mole fraction in the liquid phase.
Step-by-step explanation:
The mole fraction of ethanol in the vapor above a liquid mixture can be determined by first calculating the mole fraction of ethanol in the liquid phase and then using Raoult's law, which states that the partial vapor pressure of a component in a mixture is equal to the mole fraction of the component in the liquid phase multiplied by the vapor pressure of the pure component at the same temperature.
Without information about the vapor pressure of pure ethanol at 70°C, the mole fraction of ethanol in the vapor phase cannot be provided; it must be determined using the mole fraction in the liquid phase and the vapor pressure of pure ethanol.
In the case of the question, the student provides the mole fraction of ethanol in the liquid phase as 70% (0.70 when converted to a unitless ratio), but we also need the vapor pressure of pure ethanol at 70°C to calculate the vapor phase composition using Raoult's law. For example, if we knew that the vapor pressure of pure ethanol at 70°C was X atm, then we could apply Raoult's law which states that the partial pressure of ethanol in the vapor phase (Pethanol) is given by Pethanol = mole fraction of ethanol in liquid × vapor pressure of pure ethanol. To find the mole fraction in the vapor phase, we would divide Pethanol by the total pressure above the solution (in this case, 2 atm), assuming the vapor behaves ideally. It is important to note that the mole fraction of a component in the vapor phase is generally higher for the component with the lower boiling point due to its higher volatility.