A) \[2\times {{10}^{23}}\]
B) \[1\times {{10}^{26}}\]
C) \[1\times {{10}^{25}}\]
D) \[5\times {{10}^{27}}\]
Correct Answer: C
Solution :
Given, edge of the unit cell \[\text{= 2}\text{.0}\overset{\text{o}}{\mathop{\text{A}}}\,\] \[=2.0\times {{10}^{-8}}\,cm\] \[\therefore \] \[{{a}^{3}}=8\times {{10}^{-24}}cm\] Density of crystal, \[d=\frac{Z\times M}{{{a}^{3}}.{{N}_{0}}}\] (Where, Z = number of atoms per unit cell = 4 for fee, M = molecular mass) \[\Rightarrow \]` \[10=\frac{4\times M}{(8\times {{10}^{-24}})(6.023\times {{10}^{23}})}\] \[M=\frac{10\times 8\times 6.023}{10\times 4}\] \[=12.046\] Number of atoms of solid in 200 g \[=\frac{200}{12.046}\times 6.023\times {{10}^{23}}\] \[=16.60\times 6.023\times {{10}^{23}}=1\times {{10}^{25}}\]
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