Answer :

(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_{A}^{0} x_{A} and P_{B} < P^{0}_{B} x_{B} , 9 from Rault’s law ( P= P^{0}x)

where , P_{A} and P_{B} are the partial vapour pressure of corresponding volatile liquids in the solution whereas P_{A}^{0} and P^{0}_{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_{A}^{0} x_{A} + P^{0}_{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 : Na_{3}PO_{4}>KCl> CH_{3}OH.

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 Na_{3}PO_{4} it is a (3 Na^{3+} ,and 1 PO_{4}^{3-}) and

• In case of non-electrolyte CH_{3}OH i = 1.

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