JEE Main · 2020 · Shift-IImediumSOL-101

A set of solutions is prepared using 180 g of water as a solvent and 10 g of different non-volatile solutes A, B and C.…

Solutions · Class 12 · JEE Main Previous Year Question

Question

A set of solutions is prepared using 180 g of water as a solvent and 10 g of different non-volatile solutes A, B and C. The relative lowering of vapour pressure in the presence of these solutes are in the order

[Given, molar mass of A = 100 g mol⁻¹; B = 200 g mol⁻¹; C = 10,000 g mol⁻¹]

Options
  1. a

    B > C > A

  2. b

    C > B > A

  3. c

    A > B > C

  4. d

    A > C > B

Correct Answerc

A > B > C

Detailed Solution

Strategy:\n> Relative lowering of vapour pressure (RLVP) is a colligative property defined by xtextsolute=fracntextsolutentexttotalx_{\\text{solute}} = \\frac{n_{\\text{solute}}}{n_{\\text{total}}}. Since the mass of solute and solvent are the same, RLVP is inversely proportional to the molar mass of the solute.\n\nStep 1: Compare moles of each solute\nCommon mass = 10 g.\n- nA=10/100=0.10textmoln_A = 10 / 100 = 0.10\\text{ mol}\n- nB=10/200=0.05textmoln_B = 10 / 200 = 0.05\\text{ mol}\n- nC=10/10000=0.001textmoln_C = 10 / 10000 = 0.001\\text{ mol}\n\nStep 2: Relate moles to RLVP\nFor a fixed amount of solvent (180 g water = 10 mol), higher moles of solute result in a higher mole fraction, and thus higher RLVP.\nOrder: nA>nB>nC    textRLVPA>textRLVPB>textRLVPCn_A > n_B > n_C \implies \\text{RLVP}_A > \\text{RLVP}_B > \\text{RLVP}_C.\n\ntextAnswer:(3)\boxed{\\text{Answer: (3)}}

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