Which one of the following will have largest number of atoms?
(i) 1 g Au (s)
(ii) 1 g Na (s)
(iii) 1 g Li (s)
(iv) 1 g of Cl2 (g)
(i) Atomic Mass of Gold [Au] = 197g
197g of Gold [Au] contains 6.023×1023 atoms
Therefore 1 g of Gold will contain = [6.023×1023]/197
= 3.057×1021 atoms
(ii) Atomic Mass of Sodium [Na] = 23g
23g of Sodium [Na] contains 6.023×1023 atoms
Therefore 1 g of Sodium [Na] will contain = [6.023×1023]/23
= 2.619×1022 atoms
(iii) Atomic Mass of Lithium [Li] = 7g
7g of Lithium [Li] contains 6.023×1023 atoms
Therefore 1 g of Lithium [Li] will contain = [6.023×1023]/7
= 8.604×1022 atoms
(iv) Atomic Mass of Chlorine [Cl2] = 71g
71g of Chlorine [Cl2] contains 6.023×1023 atoms
Therefore 1 g of Chlorine [Cl2] will contain = [6.023×1023]/71
= 8.483×1021 atoms
Therefore, 1g of Lithium [Li] will contains the maximum number of atoms.