A) \[\text{1 g Au}\]
B) \[\text{1 g Na}\]
C) \[\text{1 g Li}\]
D) \[\operatorname{l}\,g\,C{{l}_{2}}\]
Correct Answer: C
Solution :
(i) \[1\,g\,Au=\frac{1}{197}\]mol atom of Au \[=\frac{1}{197}\times 6.022\times {{10}^{23}}\]atoms of Au (ii)\[1\,g\,Na=\frac{1}{23}\] atom of Na \[=\frac{1}{23}\times 6.022\times {{10}^{23}}\]atom of Na (iii) \[\text{1}\,\text{g}\,\text{Li}\,\text{=}\frac{\text{1}}{\text{7}}\]mol atom of Li \[=\frac{1}{7}\times 6.022\times {{10}^{23}}\]atom of Li (iv) \[1\,g\,C{{l}_{2}}=\frac{1}{71}\]mol molecules of \[C{{l}_{2}}\] \[=\frac{1}{71}\times 6.022\times {{10}^{23}}\]molecules of \[C{{l}_{2}}\] \[=\frac{2}{71}\times 6.022\times {{10}^{23}}\]atoms of \[C{{l}_{2}}\] Hence, 1 g lithium has the largest number of atoms.
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