JEE Main · 2019 · Shift-ImediumSOL-103

The osmotic pressure of a dilute solution of an ionic compound XY in water is four times that of a solution 0.01 M…

Solutions · Class 12 · JEE Main Previous Year Question

Question

The osmotic pressure of a dilute solution of an ionic compound XY in water is four times that of a solution 0.01 M BaCl2\mathrm{BaCl_2} in water. Assuming complete dissociation of the given ionic compounds in water, the concentration of XY (in mol L⁻¹) in solution is

Options
  1. a

    4×1024 \times 10^{-2}

  2. b

    16×10416 \times 10^{-4}

  3. c

    4×1044 \times 10^{-4}

  4. d

    6×1026 \times 10^{-2}

Correct Answerd

6×1026 \times 10^{-2}

Detailed Solution

Strategy:\n> Relate the osmotic pressures (π\pi) using the effective concentrations (iCi \cdot C). Assume complete dissociation to determine the van't Hoff factor (ii).\n\nStep 1: Determine effective concentration for } \ce{BaCl2} \text{\nFormula is \ceBaCl2\ce{BaCl2}, so n=3n = 3 ions (i=3i=3).\ni1C1=3times0.01=0.03textMi_1 C_1 = 3 \\times 0.01 = 0.03\\text{ M}\n\nStep 2: Determine effective concentration for XY\nWe are told πXY=4timesπ\ceBaCl2\pi_{XY} = 4 \\times \pi_{\ce{BaCl2}}.\niXYCXY=4times0.03=0.12textMi_{XY} \cdot C_{XY} = 4 \\times 0.03 = 0.12\\text{ M}\n\nStep 3: Solve for } C_{XY} \text{\nFor a 1:1 electrolyte \ceXY\ce{XY}, i=2i=2.\n2timesCXY=0.12    CXY=0.06textmol/L=6times102textmol/L2 \\times C_{XY} = 0.12 \implies C_{XY} = 0.06\\text{ mol/L} = 6 \\times 10^{-2}\\text{ mol/L}\n\ntextAnswer:(4)\boxed{\\text{Answer: (4)}}

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