A graph of vapour pressure and temperature for three different liquids X, Y, and Z is shown below :
The following inferences are made :
(A) X has higher intermolecular interactions compared to Y.
(B) X has lower intermolecular interactions compared to Y.
(C) Z has lower intermolecular interactions compared to Y.
The correct inference (s) is / are :

Solution

Chemistry ma’am · 4 steps

The curve that climbs first belongs to the liquid whose molecules hold each other most weakly.

1. Read the graph. X builds up vapour pressure starting from about 293 K, Y from about 313 K, and Z from about 333 K. So at any fixed temperature, X has the highest vapour pressure and Z the lowest.
2. Link to forces. Molecules escape into the vapour easily only when the attractions holding them in the liquid are weak. So a high vapour pressure at a low temperature means weak intermolecular interactions. The same liquid also reaches 760 mm Hg, its boiling point, first.

   Liquid	Vapour pressure starts rising near	Volatility	Intermolecular interactions
   X	293 K	highest	weakest
   Y	313 K	middle	middle
   Z	333 K	lowest	strongest

3. Check each inference. The order of interaction strength is X < Y < Z. So (B) is correct, (A) has it backwards, and (C) is wrong because Z has stronger interactions than Y, not weaker.
4. So. Only inference (B) is correct, which is option (a).

The idea to keep. Higher vapour pressure at a given temperature means a more volatile liquid, a lower boiling point and weaker intermolecular forces. The curve furthest to the left is the weakest-bound liquid.

Answer from the NTA final answer key. Solution by PrepWiser.