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JEE Main 201911 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
11 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Solid State

JEE Main 2019 · 11 January · Morning Shift · Chemistry · Solid State

A solid having density of 9 × 10^ 3 ~kg ~m^ -3 forms face centred...

A solid having density of 9×103 kg m−39 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3} forms face centred cubic crystals of edge length 2002pm.200 \sqrt{2} \mathrm{pm} . What is the molar mass of the solid? [Avogadro constant ≈6×1023 mol−1,π≈3]\left.\approx 6 \times 10^{23} \mathrm{~mol}^{-1}, \pi \approx 3\right]

Options

  1. A0.0432 kg mol−10.0432 \mathrm{~kg} \mathrm{~mol}^{-1}
  2. B0.0216 kg mol−10.0216 \mathrm{~kg} \mathrm{~mol}^{-1}
  3. C0.0305 kg mol−10.0305 \mathrm{~kg} \mathrm{~mol}^{-1}
  4. D0.4320 kg mol−10.4320 \mathrm{~kg} \mathrm{~mol}^{-1}

Correct answer

C. 0.0305 kg mol−10.0305 \mathrm{~kg} \mathrm{~mol}^{-1}

Step-by-step solution

Density =Z×MNA×a3=\frac{Z \times M}{N_{A} \times a^{3}} 9×103=4×M(200×2×10−12)36×10239 \times 10^{3}=\frac{4 \times \mathrm{M}}{\left(200 \times \sqrt{2} \times 10^{-12}\right)^{3} 6 \times 10^{23}}
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