JEE Main · 2020 · Shift-ImediumSOL-065

A graph of vapour pressure and temperature for three different liquids X, Y and Z is shown below: The following…

Solutions · Class 12 · JEE Main Previous Year Question

Question

A graph of vapour pressure and temperature for three different liquids X, Y and Z is shown below: image 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 inferences is/are:

Options
  1. a

    (A) and (C)

  2. b

    (A)

  3. c

    (B)

  4. d

    (C)

Correct Answera

(A) and (C)

Detailed Solution

Strategy:\n> Vapour pressure is an indicator of intermolecular forces. Liquids with stronger attractions hold their molecules more tightly and thus have a lower vapour pressure at a given temperature.\n\nStep 1: Observe the graph\nLook at the vapour pressures of X, Y, and Z at a constant temperature.\n- The curve for X is the lowest (lowest PP).\n- The curve for Z is the highest (highest PP).\n- Order of vapour pressure: PZ>PY>PXP_Z > P_Y > P_X.\n\nStep 2: Relate Vapour Pressure to Intermolecular Interactions\nStrength of interaction index is inverse to vapour pressure:\n- textStrengthX>textStrengthY>textStrengthZ\\text{Strength}_X > \\text{Strength}_Y > \\text{Strength}_Z.\n\nStep 3: Evaluate inferences\n- (A) X has higher interactions than Y: Correct, because PX<PYP_X < P_Y.\n- (B) X has lower interactions than Y: Incorrect.\n- (C) Z has lower interactions than Y: Correct, because PZ>PYP_Z > P_Y.\n\nConclusion: (A) and (C) are correct.\n\ntextAnswer:(1)\boxed{\\text{Answer: (1)}}

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A graph of vapour pressure and temperature for three different liquids X, Y and Z is shown below:… (JEE Main 2020) | Canvas Classes