A) \[4\times 6.023\times {{10}^{23}}\]
B) \[6.023\times {{10}^{23}}\]
C) \[4.25\times 6.023\times {{10}^{23}}\]
D) \[1.7\times {{10}^{24}}\]
Correct Answer: B
Solution :
As 17 g of \[N{{H}_{3}}=1\]mole of \[N{{H}_{3}}\] \[=6.023\times {{10}^{23}}\]molecules of \[N{{H}_{3}}\] \[=4\times 6.023\times {{10}^{23}}\,\]atoms \[(1\,N+3{{H}_{4}})\] 17 g of \[N{{H}_{3}}\]has \[4\times 6.023\times {{10}^{23}}\]atoms. So. 4.25 g \[N{{H}_{3}}\]will have \[=\frac{4\times 6.023\times {{10}^{23}}\times 4.25}{17}\] \[=6.023\times {{10}^{23}}\,\text{atoms}\text{.}\] Hence, the correct option is (b).
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