A) \[8.25\text{ }g/c{{m}^{3}}\]
B) \[4.25\text{ }g/c{{m}^{3}}\]
C) \[~42.5\text{ }g/c{{m}^{3}}\]
D) \[0.425\text{ }g/c{{m}^{3}}\]
Correct Answer: B
Solution :
Z for CsBr = 1 (\[\because \]one effective molecule is present) one is Cs atom and one is Br atom. \[d=\frac{ZM}{{{N}_{o}}{{a}^{3}}}\] \[=\frac{1\times 213}{{{(4.366\times {{10}^{-8}})}^{3}}\times 6.02\times {{10}^{23}}}=4.25\,g/c{{m}^{3}}\]
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