Which of the following will have largest no. of atoms?
(1) 1 g Au (g)
(2) 1g Na (s)
(3) 1g Li (s)

Explanation

We know, that number of atoms of an element
=number of moles`xxN_(A)xx`atomicity
`=("mass of the element (g)")/("gram-atomic mass of element")=xxN_(A)xx`atomicity
(`because`Gram-atomic mass of Au=197)
(2) No. of atoms present in 1 g of Na`=(1)/(23)xx6.022xx10^(23)`
(3) No. of atoms present in 1g of `Li=(1)/(7)xx6.022xx10^(23)`.