A) \[{{10}^{23}}\]
B) \[2\times {{10}^{23}}\]
C) \[6\times {{10}^{13}}\]
D) \[4\times {{10}^{23}}\]
Correct Answer: B
Solution :
No. of atoms \[=\] No. of gram molecular weight \[\times 6.02\times {{10}^{23}}\times \]atomicity No. of atoms in\[1.25\,\,g\]of\[N{{H}_{3}}\] \[=\frac{1.25}{17}\times 6.02\times {{10}^{23}}\times 4\] \[=2.2\times {{10}^{23}}\]
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