Which one of the following will have the 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)

Explanation

(i) `1g Au = 1/197 "mole" = 1/197 xx 6.022 xx 10^(23) = 3.057 xx 10^(21)` atoms
(ii) `1 g Na = 1/23 "mole" =1/23 xx 6.022 xx 10^(23) = 2.618 xx 10^(22)` atoms
(iii) `1g of Li =1/7 "mole" =1/7 xx 6.022 xx 10^(23) = 8.0603 xx 10^(22)` atoms
(iv) `1g Cl_(2) = 1/71 "mole" =1/71 xx 6.022 xx 10^(23)` molecules = `1/71 xx 6.022 xx 10^(23) xx 2` atoms
`=1.696 xx 10^(22)` atoms.
Hence, 1 g Li will contain the largest number of atoms.