JEE Main · 2019 · Shift-ImediumMOLE-103

5 moles of AB2 weigh 125 10-3\,kg and 10 moles of A2B2 weigh 300 10-3\,kg. The molar masses MA and MB in kg mol-1 are:

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

Question

55 moles of \ceAB2\ce{AB2} weigh 125×10−3 kg125 \times 10^{-3}\,\text{kg} and 1010 moles of \ceA2B2\ce{A2B2} weigh 300×10−3 kg300 \times 10^{-3}\,\text{kg}. The molar masses MAM_A and MBM_B in kg mol−1\text{kg mol}^{-1} are:

Options
  1. a

    MA=5×10−3M_A = 5 \times 10^{-3}, MB=10×10−3M_B = 10 \times 10^{-3}

    ✓
  2. b

    MA=25×10−3M_A = 25 \times 10^{-3}, MB=50×10−3M_B = 50 \times 10^{-3}

  3. c

    MA=50×10−3M_A = 50 \times 10^{-3}, MB=25×10−3M_B = 25 \times 10^{-3}

  4. d

    MA=10×10−3M_A = 10 \times 10^{-3}, MB=5×10−3M_B = 5 \times 10^{-3}

Correct Answera

MA=5×10−3M_A = 5 \times 10^{-3}, MB=10×10−3M_B = 10 \times 10^{-3}

Detailed Solution

🧠 Two equations, two unknowns Find the molar mass of each compound, set up two equations in MAM_A and MBM_B, solve.

🗺️ Set up and subtract Molar mass of \ceAB2=125×10−3/5=25×10−3 kg/mol\ce{AB2} = 125 \times 10^{-3} / 5 = 25 \times 10^{-3}\,\text{kg/mol}. Molar mass of \ceA2B2=300×10−3/10=30×10−3 kg/mol\ce{A2B2} = 300 \times 10^{-3} / 10 = 30 \times 10^{-3}\,\text{kg/mol}.

So: MA+2MB=25×10−3,2MA+2MB=30×10−3.M_A + 2M_B = 25 \times 10^{-3}, \quad 2M_A + 2M_B = 30 \times 10^{-3}.

Subtract: MA=5×10−3M_A = 5 \times 10^{-3}. Then 2MB=20×10−3  ⟹  MB=10×10−32M_B = 20 \times 10^{-3} \implies M_B = 10 \times 10^{-3}.

⚡ 5-second check The difference between \ceA2B2\ce{A2B2} and \ceAB2\ce{AB2} is exactly one extra A. So MA=30−25=5M_A = 30 - 25 = 5. The rest follows.

Answer: (a)\boxed{\text{Answer: (a)}}

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