 The figure shows a non-ideal solution with **positive deviation** from Raoult's law. For such a solution:
A$\Delta H_{\text{mix}} > 0$ and $\Delta V_{\text{mix}} < 0$
B$\Delta H_{\text{mix}} = 0$ and $\Delta V_{\text{mix}} = 0$
C$\Delta H_{\text{mix}} < 0$ and $\Delta V_{\text{mix}} < 0$
D$\Delta H_{\text{mix}} > 0$ and $\Delta V_{\text{mix}} > 0$
Answer & Solution
Correct answer: D. $\Delta H_{\text{mix}} > 0$ and $\Delta V_{\text{mix}} > 0$
Positive deviation means $p_{\text{total}} > p_A^* \chi_A + p_B^* \chi_B$. The A-B interactions are **weaker** than the A-A and B-B interactions, so molecules escape into the vapour more easily. Energy is required to push apart the original A-A and B-B contacts to form the weaker A-B contacts (endothermic mixing, $\Delta H_{\text{mix}} > 0$), and the volume of the mixture is slightly larger than the sum of pure-component volumes ($\Delta V_{\text{mix}} > 0$).
Negative deviation (option B) is the reverse: stronger A-B interactions, exothermic mixing, volume contraction. Example pairs: ethanol + water shows positive deviation; chloroform + acetone shows negative deviation.
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