Which one of the following will have largest number of atoms?
(i) 1 g of Au (s) (ii) 1 g of Na (s) (iii) 1 g of Li (s) (iv) 1g `Cl_(2) (g)`

Explanation

(i) Molar mass of Au = 197 g `mol^(-1)`
No. of atoms in 1 g of Au = `1/197 xx 6.023 xx 10^(23)`
(ii) Molar mass of Na = 23 g `mol^(-1)`
No. of atoms in 1 g of Na = `1/23 xx 6.023 xx 10^(23)`
(iii) Molar mass of Li = 7 g `mol^(-1)`
No. of atoms in 1 g of Li =` 1/7 xx 6.023 x 10^(23)`
(iv) Molar mass of CL, = 71 g `mol^(-1)`
No. of atoms in 1 g of `Cl_(2)= 1/71 xx 6.023 x 10^(23)`
Comparing the number of atoms, the largest number of atoms will be present in 1 g of Li. Since the mass is same in each case, the element with the lowest molar mass would have the largest number of atoms.
`:.` Li with lowest molar mass would have the largest number of atoms.