(i) What type of deviation from Raoult’s law is observed, when two volatile liquids A and

B on mixing produce a warm solution? Explain with the help of a well labelled vapour pressure graph.

ii) Consider separate solutions of 0.5 M CH3OH, 0.250 M KCl (aq) and 0.125 M Na3PO4 (aq). Arrange the above solutions in the increasing order of their Van’t Hoff factor.

(i) When two volatile liquids A and B on mixing produce a warm solution , heat will be released and an exothermic reaction will take place which produces negative deviation from Raoult’s law. This means that the heat released on the formation of bon ds b/w A and B is more than heat absorbed for the breakage of A-A and B-B bonds b/w molecules.

• Therefor the vapour pressure of the solution will be reduced and a negative deviation occurs.

From the labelled vapour pressure vs mole fraction graph we can say that

• Negative Deviation occurs when the total vapour pressure of the solution (non-ideal ) is less than what it should be according to Raoult’s Law (ideal solution) .

• Considering the same A and B components to form a non-ideal solution, it will show negative deviation from Raoult’s Law only when:

• P_A < P_A⁰ x_A and P_B < P⁰_B x_B , 9 from Rault’s law ( P= P⁰x)

where , P_A and P_B are the partial vapour pressure of corresponding volatile liquids in the solution whereas P_A⁰ and P⁰_B are the vapour pressure of the pure liquids and x_A and x_B are the mole fractions of liquids A and B in the soln.

• as the total vapour pressure (P_A⁰ x_A + P⁰_B x_B) is less than what it should be with respect to Raoult’s Law.

• The liquid-liquid interaction among the 2 liquids is stronger than the liquid-liquid between themselves interaction that is, A – B > A – A or B – B.

• The enthalpy of mixing is negative, Δ_mix H < 0 (exothermic reaction) because more heat is released when new molecular interactions are formed b/w A and B and the volume of mixing is negative also, Δ_mix V < 0 as the volume decreases on the dissolution of volatile liquid components A and B.

(ii)

• The correct increasing order of their Van’t Hoff factor is : Na₃PO₄>KCl> CH₃OH.

Van’t Hoff factor is designated as,

• Hence, for the given ionic solutions i will be equal to the total number of ions dissociated in the solutions.

• In case of KCl , no. Of ions in the solution is 2 (K⁺ and Cl^- )

• Whereas in case of Na₃PO₄ it is a (3 Na³⁺ ,and 1 PO₄^3-) and

• In case of non-electrolyte CH₃OH i = 1.