7. Determine the density of cesium chloride which crystallizes in a bcc structre with edge
length of unit cell 4.12 x 10cm
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Explanation:
strcuture of Cesium chloride is body centre cubic type in which Cs+ placed within centre and Cl- placed at corner of lattice.
number of CI- ions per unit cell = 8 x 1/8 = = 1 1
number of Cs+ ions per unit cell = 1 so, there is only one molecule of CsCl in
a lattice .
so, mass of one molecule of CsCl = molar mass of CsCI/Avogadro's number
=
(133 + 35.5)/(6.023 x 1023)
168.5/6.023 × 1023
= 2.8 x 10^-22 g
volume of lattice = a° , where a is edge =
length
= (412.1 × 10^-12)
= 7 10^-23 cm 3
so, density of CsCl crystal = 2.8 x
10^-22/7 x 10^-23
= 4 g/cm
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