A) \[6.023\times 1023\]
B) \[4\times 6.023\times 1023\]
C) \[1.7\times 1024\]
D) \[4.25\times 6.023\times 1023\]
Correct Answer: A
Solution :
Number of atoms in 4.25 g of \[\text{N}{{\text{H}}_{\text{3}}}\] \[=\frac{4.25}{17}\times {{N}_{A}}\times 4\] \[={{N}_{A}}=6.023\times {{10}^{23}}\]
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