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

Respuesta :

Answer:

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 [tex]P_A[/tex] and vapour pressure of component B is [tex] P_B[/tex].

[tex] P^0_A[/tex]= Vapour pressure of component A in pure form

[tex] P^0_B[/tex]= Vapour pressure of component B in pure form

[tex] x_A[/tex]=Mole fraction of component A

[tex] x_B=[/tex]=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.

[tex]P_A >P^0_A\cdot x_A[/tex],[tex]P_B>P^0_B\cdot x_B[/tex]

Therefore, [tex]P_A+P_B >P^0_A\cdot x_A+P^0_B \cdot x_B[/tex]

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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