JEE Main · 2019 · Shift-ImediumSOL-009

The mole fraction of a solvent in aqueous solution of a solute is 0.8. The molality (in mol kg⁻¹) of the aqueous…

Solutions · Class 12 · JEE Main Previous Year Question

Question

The mole fraction of a solvent in aqueous solution of a solute is 0.8. The molality (in mol kg⁻¹) of the aqueous solution is:

Options
  1. a

    13.88×10−313.88 \times 10^{-3}

  2. b

    13.88×10−113.88 \times 10^{-1}

  3. c

    13.8813.88

    ✓
  4. d

    13.88×10−213.88 \times 10^{-2}

Correct Answerc

13.8813.88

Detailed Solution

Strategy:\n> Relate mole fraction (xx) to molality (mm) using the molar mass of the solvent (water = 18). m=fracxtextsolutextextsolvent⋅Mtextsolvent(kg)m = \\frac{x_{\\text{solute}}}{x_{\\text{solvent}} \cdot M_{\\text{solvent(kg)}}}.\n\nAnalysis:\n- xtextsolvent=0.8x_{\\text{solvent}} = 0.8.\n- xtextsolute=1−0.8=0.2x_{\\text{solute}} = 1 - 0.8 = 0.2.\nm=frac0.20.8times0.018=frac0.20.0144≈13.88textmm = \\frac{0.2}{0.8 \\times 0.018} = \\frac{0.2}{0.0144} \approx 13.88\\text{ m}\n\ntextAnswer:(3)\boxed{\\text{Answer: (3)}}

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