A) \[1\times {{10}^{25}}\]
B) \[1\times {{10}^{24}}\]
C) \[1\times {{10}^{22}}\]
D) \[1\times {{10}^{20}}\]
Correct Answer: A
Solution :
| [a] Edge length of the unit cell |
| \[=2{{A}^{0}}=2\times {{10}^{-8}}cm\] |
| Volume of the unit cell \[={{(2\times {{10}^{-8}})}^{3}}c{{m}^{3}}\] |
| \[=8\times {{10}^{-24}}c{{m}^{3}}\] |
| Mass of unit cell\[\text{= volume }\!\!\times\!\!\text{ density}\] |
| \[=8\times {{10}^{-24}}\times 2.5g\] |
| No. of unit cells in 200 g of the metal |
| \[\frac{\text{mass of metal}}{\text{mass of unitcell}}=\frac{200}{8\times {{10}^{-24}}\times 2.5}=1\times {{10}^{25}}\] |
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