JEE Main · 2022 · Shift-ImediumMOLE-108

A commercially sold conc. HCl is 35\% HCl by mass. If the density of this commercial acid is 1.46\,g/mL, the molarity…

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

Question

A commercially sold conc. HCl is 35%35\% HCl by mass. If the density of this commercial acid is 1.46g/mL1.46\,\text{g/mL}, the molarity of this solution is: (Atomic mass: Cl =35.5= 35.5, H =1amu= 1\,\text{amu})

Options
  1. a

    10.2M10.2\,\text{M}

  2. b

    14.0M14.0\,\text{M}

  3. c

    12.5M12.5\,\text{M}

  4. d

    18.2M18.2\,\text{M}

Correct Answerb

14.0M14.0\,\text{M}

Detailed Solution

🧠 The mass-percent-to-molarity formula M=10×d×%Mw.M = \frac{10 \times d \times \%}{M_w}.

This compresses the volume-and-mass conversion into one line.

🗺️ Plug in d=1.46g/mLd = 1.46\,\text{g/mL}, %=35\% = 35, Mw(\ceHCl)=36.5g/molM_w(\ce{HCl}) = 36.5\,\text{g/mol}.

M=10×1.46×3536.5=51136.514.0M.M = \frac{10 \times 1.46 \times 35}{36.5} = \frac{511}{36.5} \approx 14.0\,\text{M}.

Quick mental math 1.46×25=36.51.46 \times 25 = 36.5. So 1.46/36.5=1/25=0.041.46 / 36.5 = 1/25 = 0.04. Then M=10×0.04×35=14M = 10 \times 0.04 \times 35 = 14.

Answer: (b)\boxed{\text{Answer: (b)}}

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