Question

# All of the below given compounds forms ideal solutions except

A

${\mathrm{C}}_{2}{\mathrm{H}}_{6}\mathrm{and}{\mathrm{C}}_{2}{\mathrm{H}}_{5}\mathrm{I}$

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B

${\mathrm{C}}_{2}{\mathrm{H}}_{5}\mathrm{I}\mathrm{and}{\mathrm{C}}_{2}{\mathrm{H}}_{5}\mathrm{OH}$

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C

${\mathrm{C}}_{6}{\mathrm{H}}_{6}\mathrm{and}{\mathrm{C}}_{6}{\mathrm{H}}_{5}{\mathrm{CH}}_{3}$

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D

${\mathrm{C}}_{6}{\mathrm{H}}_{5}\mathrm{Cl}\mathrm{and}{\mathrm{C}}_{6}{\mathrm{H}}_{5}\mathrm{Br}$

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Solution

## The correct option is B ${\mathrm{C}}_{2}{\mathrm{H}}_{5}\mathrm{I}\mathrm{and}{\mathrm{C}}_{2}{\mathrm{H}}_{5}\mathrm{OH}$Liquid-liquid solutions can be classified into ideal and non-ideal solutions on the basis of Raoult’s law: Raoult's law: States that for any solution, the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction.When the solute is non-volatile, only the solvent molecules are present in the vapour phase and contribute to vapour pressure. Let ${\mathrm{p}}_{1}$ be the vapour pressure of the solvent, ${\mathrm{x}}_{1}$ be its mole fraction, and ${{\mathrm{p}}_{1}}^{\mathrm{o}}$ be the vapour pressure of the pure solvent. Then according to Raoult’s law ${\mathrm{p}}_{1}\propto {\mathrm{x}}_{1}\phantom{\rule{0ex}{0ex}}{\mathrm{p}}_{1}={{\mathrm{p}}_{1}}^{\mathrm{o}}{\mathrm{x}}_{1}\phantom{\rule{0ex}{0ex}}$The explanation for the correct option:Non-ideal solutions: The solutions which do not obey Raoult's law over the entire range of concentration are known as non-ideal solutions. The vapour pressure of such a solution is either higher or lower than that predicted by Raoult’s law. If it is higher, the solution exhibits positive deviation and if it is lower, it exhibits negative deviation from Raoult’s law.Positive deviation - In the case of positive deviation from Raoult’s law, A-B interactions are weaker than those between A-A or B-B, i.e., in this case, the intermolecular attractive forces between the solute-solvent molecules are weaker than those between the solute-solute and solvent-solvent molecules. Therefore in such solutions, molecules of A or B will find it easier to escape than in a pure state. This will increase the vapour pressure and result in a positive deviation. Examples are the solution of ethanol and acetone, carbon disulphide and acetone, etc. Negative deviation - In the case of negative deviation from Raoult’s law, A-B interactions are stronger than those between A-A or B-B, i.e., in this case, the intermolecular attractive forces between the solute-solvent molecules are stronger than those between the solute-solute and solvent-solvent molecules. Therefore in such solutions, the molecules do not escape which leads to a decrease in vapour pressure resulting in negative deviation. Examples are the solution of phenol and aniline, chloroform and acetone, etc. The explanation for incorrect options: Ideal solutions: The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions.The ideal behaviour of the solutions can be explained by considering two components A and B. In pure components, the intermolecular attractive interactions will be of types A-A and B-B, whereas in the binary solutions in addition to these two interactions, A-B type of interactions will also be present. If the intermolecular attractive forces between the A-A and B-B are nearly equal to those between A-B, this leads to the formation of the ideal solution.Examples-. Solution of n-hexane and n-heptane, bromoethane and chloroethane, benzene and toluene, etc.Therefore, (A), (C) and (D) are incorrect options. Hence, the correct option is (B) ${\mathrm{C}}_{2}{\mathrm{H}}_{5}\mathrm{I}\mathrm{and}{\mathrm{C}}_{2}{\mathrm{H}}_{5}\mathrm{OH}$ as it is the only non-ideal solution out of the given options.

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