To determine which of the given mixtures shows a positive deviation from Raoult's law, let's understand the concept of positive deviation.
Raoult’s Law: It states that the partial vapor pressure of each component in a solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution. Mathematically, for a binary solution:
\(P_A = P_A^0 \cdot x_A\) and \(P_B = P_B^0 \cdot x_B\)
Where \(P_A\) and \(P_B\) are the partial vapor pressures, \(P_A^0\) and \(P_B^0\) are the vapor pressures of pure components A and B, and \(x_A\) and \(x_B\) are the mole fractions of components A and B, respectively.
Positive Deviation: A solution exhibits positive deviation from Raoult's law if the interactions between the different molecules in the mixture are weaker than the interactions in the pure components. As a result, the vapor pressure of the mixture is higher than expected.
In case of positive deviation, the mixture has higher vapor pressure and lower boiling point compared to ideal solutions. This occurs because component molecules prefer to interact with their own kind rather than the other component, leading to weaker intermolecular forces in the mixture.
Conclusion: Based on the interactions between the molecules, the correct answer is that the mixture Ethanol + Acetone exhibits positive deviation from Raoult's law.