A) \[\text{6}.0\text{2 }\times \text{l}{{0}^{\text{23}}}\text{ atoms of H}\]
B) 4 g atoms of Na
C) \[\text{6}.0\text{2 }\times \text{ 1}{{0}^{\text{23}}}\text{ atoms of Na}\]
D) \[\text{4 moles of NaOH}\]
Correct Answer: C
Solution :
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