A) \[2.57\times {{10}^{21}}\]
B) \[5.14\times {{10}^{21}}\]
C) \[1.28\times {{10}^{21}}\]
D) \[1.71\times {{10}^{21}}\]
Correct Answer: A
Solution :
Mass of one unit-cell\[(m)=\]volume\[\times \]density \[={{a}^{3}}\times d\] \[={{a}^{3}}\times \frac{MZ}{{{N}_{0}}{{a}^{3}}}=\frac{MZ}{{{N}_{0}}}\] \[m=\frac{58.5\times 4}{6.02\times {{10}^{23}}}g\] \[\therefore \]Number of unit cells in\[1\,\,g=\frac{1}{m}\] \[=\frac{6.02\times {{10}^{23}}}{58.5\times 4}=2.57\times {{10}^{21}}\]
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