JEE Main · 2024 · Shift-ImediumMOLE-132

The density of "x M" solution of NaOH is 1.12\,g mL-1, while in molality, the concentration of the solution is 3\,m (3…

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

Question

The density of "xx M" solution of NaOH is 1.12 g mL−11.12\,\text{g mL}^{-1}, while in molality, the concentration of the solution is 3 m3\,m (3 molal). Then xx is: (Molar mass of NaOH =40 g/mol= 40\,\text{g/mol})

Options
  1. a

    3.53.5

  2. b

    3.83.8

  3. c

    2.82.8

  4. d

    3.03.0

    ✓
Correct Answerd

3.03.0

Detailed Solution

🧠 The standard MM–mm–dd formula To switch between molarity and molality with density, use: M=1000⋅m⋅d1000+m⋅Msolute.M = \frac{1000 \cdot m \cdot d}{1000 + m \cdot M_\text{solute}}.

🗺️ Plug and play m=3m = 3, d=1.12d = 1.12, Msolute=40M_\text{solute} = 40. M=1000×3×1.121000+3×40=33601120=3.0 M.M = \frac{1000 \times 3 \times 1.12}{1000 + 3 \times 40} = \frac{3360}{1120} = 3.0\,\text{M}.

⚡ Speed scan The numerator 33603360 is exactly 3×11203 \times 1120. So M=3.0M = 3.0 falls out without a calculator.

Answer: (d)\boxed{\text{Answer: (d)}}

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