A) \[42.5\text{ }g/c{{m}^{3}}\]
B) \[0.425\text{ }g/c{{m}^{3}}\]
C) \[8.25\text{ }g/c{{m}^{3}}\]
D) \[4.25\text{ }g/c{{m}^{3}}\]
Correct Answer: D
Solution :
| [d] 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 × 10-10 cm |
| \[\therefore \,\,\text{Density}\,=\frac{2\times 213}{{{(436.6\times {{10}^{-10}})}^{3}}\times 6.02\times {{10}^{23}}}\] |
| = 8.50 g/cm3 |
| For a unit cell = \[\frac{8.50}{2}=4.25\,g/c{{m}^{3}}\] |
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