A) \[5\times {{10}^{24}}\]
B) \[5\times {{10}^{25}}\]
C) \[6\times {{10}^{23}}\]
D) \[2\times {{10}^{25}}\]
Correct Answer: A
Solution :
[a] \[~5\times {{10}^{24}}\] \[P=\frac{z\times M}{{{a}^{3}}\times {{10}^{-3}}\times NO}\] \[or\,M=\frac{10\times {{(200)}^{3}}\times {{10}^{-30}}\times 6\times {{10}^{23}}}{4}=12\] Thus, 12g contain \[=\text{ }No=6\times {{10}^{23}}\]atom \[\therefore \,100g\,will\,contain\,=\frac{6\times {{10}^{23}}}{12}\times 100\] \[=5\times {{10}^{24}}\]
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