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) \[\text{=}\,\,\text{volume}\,\text{ }\!\!\times\!\!\text{ }\,\text{density}\] \[={{a}^{3}}\times d={{a}^{3}}\times \frac{Mz}{{{N}_{0}}{{a}^{3}}}=\frac{Mz}{{{N}_{0}}}\] \[m=\frac{58.5\times 4}{6.02\times {{10}^{23}}}g\] \[\therefore \]Number of unit cells in \[1\,g=\frac{1}{m}\] \[=\frac{6.02\times {{10}^{23}}}{58.5\times 4}\] \[=2.57\times {{10}^{21}}\]
You need to login to perform this action.
You will be redirected in
3 sec
