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How many unit cells are present in a cube shaped ideal crystal of NaCl of mass 1.00 g ?

Options

  1. A2.57 × 10 21
  2. B5.14 × 10 21
  3. C1.28 × 10 21
  4. D1.71 × 10 21

Correct answer

A. 2.57 × 10 21

Step-by-step solution

∴ Mass of one unit cell = V × d (V is volume, disdensity) = a 3 × d A l s o , d e n s i t y = n × a t . w t . a 3 × a v . no. ∴ Mass of one unit cell = a 3 × n × at .wt . a 3 × av .no . = n × at .wt . av .no . = 4×58 .5 6 .02×10 23 (n = 4 for cubic shape) = 38.87 × 10 - 23 g No .   of   unit   cell   in   1   g   =   1 38 .87   ×   10 23   =   2 .57   ×   10 21

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