Q. 125.0( 1 Vote )
(i) What type of
(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:
• PA < PA0 xA and PB < P0B xB , 9 from Rault’s law ( P= P0x)
where , PA and PB are the partial vapour pressure of corresponding volatile liquids in the solution whereas PA0 and P0B are the vapour pressure of the pure liquids and xA and xB are the mole fractions of liquids A and B in the soln.
• as the total vapour pressure (PA0 xA + P0B xB) 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, Δmix 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, Δmix V < 0 as the volume decreases on the dissolution of volatile liquid components A and B.
• 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 PO43-) and
• In case of non-electrolyte CH3OH i = 1.
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