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
- A6.250 × 1 0 22
- B3.125 × 1 0 23
- C3.125 × 1 0 22
- 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