A) 1 g Au
B) 1 g Na
C) 1 g Li
D) \[1\,g\,C{{l}_{2}}\]
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) \[1\,g\,Na=\frac{1}{23}mol\] atom of Na \[=\frac{1}{23}\times 6.022\times {{10}^{23}}\] atom of Na (iii) \[1\,g\,Li=\frac{1}{7}mol\] 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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