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NTA Abhyas JEE Main2020ChemistrySolid StatePractice

A metal of density 7.5 × 1 0 3 kg m - 3 has an fcc crystal structure with lattice parameter a = 400 pm. Calculate the number of unit cells present in 0.015 kg of the metal.

Options

  1. A6.250 × 1 0 22
  2. B3.125 × 1 0 23
  3. C3.125 × 1 0 22
  4. D1.563 × 1 0 22

Correct answer

C. 3.125 × 1 0 22

Step-by-step solution

V 1 = a 3 = ( 400 × 1 0 - 12 m ) 3 = 64 × 1 0 - 30 m 3 V 2 = m a s s d e n s i t y = 0.015 k g 7.5 × 1 0 - 3 m - 3 = 2 × 1 0 - 6 m 3 where V 1 is the volume of the unit cell and V 2 that of the metal sample. So number of unit cell = T o t a l volume Volumeofunitcell = 2 × 1 0 - 6 m 3 64 × 1 0 - 30 m 3 = 3.125 × 1 0 22

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