JEE Main · 2024 · Shift-IImediumMOLE-104

Molality (m) of 3\,M aqueous solution of NaCl is: (Density = 1.25\,g mL-1; Mr: Na = 23, Cl = 35.5)

Some Basic Concepts (Mole Concept) · Class 11 · JEE Main Previous Year Question

Question

Molality (mm) of 3 M3\,\text{M} aqueous solution of NaCl is: (Density =1.25 g mL−1= 1.25\,\text{g mL}^{-1}; MrM_r: Na =23= 23, Cl =35.5= 35.5)

Options
  1. a

    1.9 m1.9\,m

  2. b

    3.85 m3.85\,m

  3. c

    2.79 m2.79\,m

    ✓
  4. d

    2.90 m2.90\,m

Correct Answerc

2.79 m2.79\,m

Detailed Solution

🧠 Subtract the salt mass to find solvent mass A 1 L1\,\text{L} of solution weighs 1250 g1250\,\text{g} (1000×1.251000 \times 1.25). Of that, 175.5 g175.5\,\text{g} is salt. The rest is water.

🗺️ One-litre sandbox Mass of \ceNaCl=3×58.5=175.5 g\ce{NaCl} = 3 \times 58.5 = 175.5\,\text{g}. Mass of solution =1000×1.25=1250 g= 1000 \times 1.25 = 1250\,\text{g}. Mass of water =1250−175.5=1074.5 g=1.0745 kg= 1250 - 175.5 = 1074.5\,\text{g} = 1.0745\,\text{kg}.

m=31.0745≈2.79 m.m = \frac{3}{1.0745} \approx 2.79\,\text{m}.

⚠️ Sanity check Density above 11 → molality always less than molarity. So the answer must be below 33. That kills options (b) and (d).

Answer: (c)\boxed{\text{Answer: (c)}}

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Molality (m) of 3\,M aqueous solution of NaCl is: (Density = 1.25\,g mL-1; Mr: Na = 23, Cl = 35.5) (JEE Main 2024) | Canvas Classes