Question -
Answer -
1 g of Au (s) 
mol of Au (s)
atoms of Au (s)
= 3.06 × 1021atoms of Au (s)
1 g of Na (s) =
mol of Na (s)
atoms of Na (s)= 0.262 × 1023 atoms of Na(s)
= 26.2 × 1021 atoms of Na(s)
1 g of Li (s) 
mol of Li (s)
atoms of Li (s)= 0.86 × 1023 atoms of Li(s)
= 86.0 × 1021 atoms of Li(s)
1 g of Cl2 (g) 
mol of Cl2 (g)(Molar mass of Cl2 molecule= 35.5 × 2 = 71 g mol–1)
molecules of Cl2 (g)
= 0.0848 × 1023 moleculesof Cl2 (g)
= 8.48 × 1021 molecules ofCl2 (g)
As one molecule of Cl2 containstwo atoms of Cl.
Number of atoms of Cl = 2× 8.48 × 1021 =16.96× 1021 atoms of Cl
Hence, 1 g of Li (s) will have the largest number ofatoms.