Positive deviation from raoult's law occurs when the vapour pressure of component is greater than what is expected in raoult's law. for example, consider two components a and b to form non-ideal solutions. let the vapour pressure, pure vapour pressure and mole fraction of component a be pa, pao and xa respectively and that of component b be ps, pb° and xb respectively. these liquids will show positive deviation when raoult's law when: a. pia]> p[ao] times x[a] and pib] > pibo] times xib], as the total vapour pressure (piao] xia] + p[b0] x[b]) is greater than what it should be according to raoult's law. o b. the solute-solvent forces of attraction is weaker than solute-solute and solvent-solvent interaction o c. the enthalpy of mixing is positive because weaker binding forces or even repulsion are resulted o d. the volume of mixing is positive as weaker binding forces have led to an expansion in volume o a and b only o a, b and c only o all of the above

All of the above.

Explanation:

In positive deviation from Raoult's  Law occur when the vapour pressure of components is greater than what is expected value in Raoult's law.

When a solution is non ideal then it shows positive or negative deviation.

Let two solutions A and B to form non- ideal solutions.let the vapour pressure  of component A is and vapour pressure of component B is .

= Vapour pressure of component A in pure form

= Vapour pressure of component B in pure form

=Mole fraction of component A

=Mole fraction of component B

The interaction between A- B is less than the interaction A- A and B-B interaction.Therefore, the escaping tendency of liquid molecules in mixture is greater than the escaping tendency in pure form.Hence, the vapour pressure of a mixture is greater than the initial value of vapour pressure.

,

Therefore,

Therefore, the enthalpy of mixing is greater than zero and change in volume is greater than zero.

Hence, option a,b,c and d are true.

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