A) \[42.5\,g/c{{m}^{3}}\]
B) \[0.425\,g/\,c{{m}^{3}}\]
C) \[8.25\,g/\,c{{m}^{3}}\]
D) \[4.25\,g/c{{m}^{3}}\]
Correct Answer: D
Solution :
Density of \[CsBr\,=\frac{Z\times m}{{{a}^{3}}\times {{N}_{0}}}\] \[Z\to \] no. of atoms in the bcc unit cell = 2 \[M\to \] molar mass of CsBr = 133 + 80 = 2 \[a\to \] edge length of unit cell = 436.6 pm \[=436.\,6\times {{10}^{-10}}\,cm\] \[\therefore \,\,Density\,=\frac{2\times 213}{{{(436.6\times {{10}^{-10}})}^{3}}\times 6.02\times {{10}^{23}}}\] \[=8.50\,g\,/c{{m}^{3}}\] For a unit cell = \[\frac{8.50}{2}=4.25\,g/c{{m}^{3}}\]
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