Which one of the following graphs accurately represents the plot of partial pressure of CS₂ vs its mole fraction in a mixture of acetone and CS₂ at constant temperature?

Solution

Chemistry ma’am · 3 steps

Acetone and CS2 attract each other weakly, so the CS2 curve bows above the Raoult line.

1. Type of deviation. Acetone is polar and CS₂ is non-polar. The acetone–CS₂ attractions are weaker than the acetone–acetone and CS₂–CS₂ attractions, so CS₂ molecules escape more easily from the mixture. This is a positive deviation: p_(CS₂) is higher than Raoult's law predicts.

   p_(CS₂) > x_(CS₂) p°_(CS₂) for 0 < x_(CS₂) < 1

2. The two ends. When CS₂ is dilute (x_(CS₂) → 0) it behaves as a solute and follows Henry's law, p = K_H x, with K_H > p°_(CS₂). So the curve starts along the steep Henry's line. When CS₂ is nearly pure (x_(CS₂) → 1) it is the solvent and follows Raoult's law, so the curve flattens and meets p°_(CS₂).

   Option	What the CS₂ curve shows	Matches acetone + CS₂?
   (a)	Above the Raoult line, Henry's law at low x, reaches p°_(CS₂)	Yes
   (b)	Below the Raoult line, total pressure has a minimum	No, that is negative deviation
   (c)	Straight lines, no deviation	No, that is an ideal solution
   (d)	CS₂ curve still below its Raoult line	No, the CS₂ curve must lie above

3. So. The partial pressure of CS₂ rises above the Raoult line, follows Henry's law at low mole fraction and joins Raoult's law near pure CS₂. That is graph (a).

The idea to keep. Weaker A–B attraction than A–A and B–B means positive deviation: each partial pressure curve sits above its Raoult line, follows Henry's law when dilute and meets Raoult's law near the pure liquid.

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