Multiple choice

An element has a body-centred cubic (BCC) structure with a cell edge of 210 pm. The density of the element is 4.2 g/cm3. Number of atoms present in 210 g of the element is

  1. 5.03 x 1024

  2. 10.06 ×1024

  3. 20.12 × 1024

  4. 30.18 ×1024

  5. 40.24 ×1024
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Volume of the unit cell = (210 pm)3= (210 ×10-12 m)3 = (210 × 10-10 cm)3 = 9.94 × 10-24 cm3 Volume of 208 g of the element = mass / density = 210 g / 4.2 g cm-3 = 50 cm3 Number of unit cells in this volume = 50 cm3 / 9.94 x 10-24 cm3 = 5.03 x 1024 Since each bcc cubic unit cell contains 2 atoms, so the total number of atoms in 210 g of the element = 2 (atoms / unit cell) × 5.03 x 1024 unit cells = 10.06×1024 atoms

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