Question

which one of the following will have largest number of atoms

Answer 1

Answer:

(i) 1 g Au = 1/197 mol = 1/197×6.022×10^23 atoms

(ii) 1 g Na = 1/23 mol = 1/23×6.022×10^23 atoms

(iii) 1 g Li = 1/7 mol = 1/7×6.022×10^23 atoms

(iv) 1 g Cl2 = 1/71 mol = 1/71×6.022×10^23 atoms

Thus, 1 g of Li has the largest number of atoms.

Explanation:

Here 6.022 ×10^23 is avogadro's no.