Question

In a mixture, ‘A’ and ‘B’ components show the negative deviations as:
A. \[\Delta {V_{mix}} > 0\]
B. \[\Delta {V_{mix}} < 0\]
C. \[A - B\] interaction is weaker than \[A - A\]and \[B - B\] interactions
D. None of the above reasons is correct

Answer 1

Hint: The interactive forces between the solvent and solute molecules in a solution play a major role in deciding the direction of deviation of the vapor pressure of the mixture from its pure state. Hence the deviation is positive or negative accordingly.

Complete answer:
Raoult’s law: It states that the partial vapor pressure of the solvent in a mixture is identical to the product of pressure of solvent in pure state and its mole fraction in the solution. The expression for Raoult’s law can be given as follows:
\[{p_A} = {\chi _A}P\]
Where, \[{p_A} \Rightarrow \] vapor pressure of component ‘A’ in the mixture
\[{\chi _A} \Rightarrow \] Mole fraction of component ‘A’ in the mixture
\[P \Rightarrow \] Pressure of pure component ‘A’
Reason for observation of deviations:
The interactive forces between the \[A - B\], \[A - A\] and \[B - B\] interactions are uniform throughout the mixture for an ideal solution. But in the case of a non-ideal solution, the size of each component varies and due to this the forces of interaction between each component also changes. Because of these differences in the strength of forces, the mixture does not obey Raoult’s law and thus, the deviations in the vapor pressure (can be positive or negative) are observed in case of non-ideal solution.
Negative deviation:
If the force of attraction between the molecules of components ‘A’ and ‘B’ in the mixture are stronger than the attractive forces between \[A - A\] and \[B - B\], then the propensity of escaping of \[AB\] molecules from the solution become less than that of pure liquids A and B. Therefore, the partial pressure of each component in the mixture will be less as compared to the expected vapor pressure of the ideal solution. Thus, negative deviations from Raoult’s law can be observed.
Conclusion: Because \[A - B\] interaction is stronger than \[A - A\]and \[B - B\] interactions i.e., the molecules in the mixture are held more tightly. Therefore, the volume of the mixture decreases and hence, \[\Delta {V_{mix}}\] will be negative i.e., \[\Delta {V_{mix}} < 0\].
So, option (B) is the correct answer.

Note:
When \[A - B\] interaction is weaker than \[A - A\]and \[B - B\] interactions, then the tendency for escaping of A and B molecules from the solution increases from that of pure molecules. Therefore, the partial pressure of each component in the mixture will be greater as compared to the expected vapor pressure of the ideal solution. Thus, show positive deviation from Raoult's law.