A binary liquid solution is prepared by mixing n - heptane and ethanol. Which one of the following statements is correct regarding the behaviour of the solution ?

  • Option 1)

    The solution formed is an ideal solution

  • Option 2)

    The solution is non-ideal, showing +Ve deviation from Raoult’s law

  • Option 3)

    The solution is non - ideal ,showing -Ve deviation from Raoult’s law

  • Option 4)

    n - heptane shows +Ve deviation while ethanol shows  -Ve deviation from Raoult’s law.

 

Answers (1)

As we learnt in

Condition for positive deviation -

When A - A interactions and B - B  interactions  are  stronger than A - B interactions.

e.g. Ethanol + water

- wherein

\Delta H_{mix}> 0

\Delta V_{mix}> 0

(V_{f}>V_{A}+V_{B} )

\Delta H - Change\: in\: enthalpy

\Delta V - Change\: in\: volume

 

 Between two n-hexane molecule the force interaction is Vander Waal and between two molecule ethanol the force of interaction is Hydrogen bond. Hence when both are mixed together their interaction are weaker than previous one so they shows positive deviation.

Correct option is 3.

 


Option 1)

The solution formed is an ideal solution

This is an incorrect option.

Option 2)

The solution is non-ideal, showing +Ve deviation from Raoult’s law

This is an incorrect option.

Option 3)

The solution is non - ideal ,showing -Ve deviation from Raoult’s law

This is an incorrect option.

Option 4)

n - heptane shows +Ve deviation while ethanol shows  -Ve deviation from Raoult’s law.

This is an incorrect option.

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