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The Solid State

Question
CBSEENCH12011391

CsBr crystallises in a body centred cubic lattice. The unit cell length is 436.6 pm. Given that the atomic mass of Cs - 133 and that of Br = 80 amu and Avogadro number being
6.02 x 1023 mol-1, the density of CsBr is:

  • 42.5 g/cm3

  • 0.425 g/cm3

  • 8.25 g/cm3

  • 4.25 g/cm3

Solution

D.

4.25 g/cm3

Density space of space CsBr space equals space fraction numerator straight Z space straight x space straight M over denominator straight a cubed space straight x space straight N subscript straight o end fraction
straight Z space rightwards arrow space no. space of space atoms space in space the space bcc space unit space cell space equals 2
straight M space rightwards arrow space molar space mass space of space CsBr space equals space 133 space plus space 80 space equals space 213
straight a space rightwards arrow space edge space length space of space unit space cell space equals space 436.6 space pm
space equals space 436.6 space straight x space 10 to the power of negative 10 end exponent space cm
Therefore comma space
Density space equals space fraction numerator 2 space straight x space 213 over denominator left parenthesis 436.6 space straight x space 10 to the power of negative 10 end exponent right parenthesis cubed space straight x space 6.02 space straight x space 10 to the power of 23 end fraction
space equals space 8.50 space straight g space divided by cm cubed
For space straight a space unit space cell space equals space fraction numerator 8.50 over denominator 2 end fraction space equals space 4.25 space straight g divided by space cm cubed