In mixture  A and B components show -ve deviation as

  • Option 1)

    \Delta V_{mix}> 0

  • Option 2)

    \Delta V_{mix}< 0

  • Option 3)

    A-B\: \: interaction\: is\: weaker\: than\: A-A\: and\: B-B\: interaction

  • Option 4)

    A-B \: interaction\: is\: stronger\: than\: A-A\: and\: B-B\: interaction

 

Answers (2)

As we learnt in

Condition for negative deviation -

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

e.g. Acetone + Chloroform

- 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

 

 

 

For negative deviation, from Raoult's law, \Delta V_{mix}< 0\: and \: \Delta H_{mix}< 0 Here A-B attractive force is greater than A - A and B - B attractive forces

Correct option is 2.


Option 1)

\Delta V_{mix}> 0

This is an incorrect option.

Option 2)

\Delta V_{mix}< 0

This is the correct option.

Option 3)

A-B\: \: interaction\: is\: weaker\: than\: A-A\: and\: B-B\: interaction

This is an incorrect option.

Option 4)

A-B \: interaction\: is\: stronger\: than\: A-A\: and\: B-B\: interaction

This is an incorrect option.

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