A) \[42.5g/c{{m}^{3}}\]
B) \[0.425g/c{{m}^{3}}\]
C) \[8.25\text{ }g/c{{m}^{3}}\]
D) \[4.25\text{ }g/c{{m}^{3}}\]
Correct Answer: D
Solution :
Density of \[CsBr=\frac{Z\times M}{{{a}^{3}}\times {{N}_{0}}}\] where \[Z=\]no. of atoms in the bcc unit cell\[=2\] \[M=\]molar mass of\[CsBr=133+80=213\] \[a=\]edge length of unit cell = 436.6 pm \[=436.6\times {{10}^{-10}}cm\] \[\therefore \]Density \[=\frac{2\times 213}{{{(436.6\times {{10}^{-10}})}^{3}}\times 6.02\times {{10}^{23}}}\] \[=8.50g/c{{m}^{3}}\] For a unit cell density\[=\frac{8.50}{2}=4.25\text{ }g/c{{m}^{3}}\]
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