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(i) 1 g Au (s)
(ii) 1 g Na (s)
(iii) 1 g Li (s)
(iv) 1 g of Cl2(g)
Asked by Briyal Last Modified
Answer 
Nurullmobeen
1 g of Au (s) 
mol of Au (s)
atoms of Au (s)
= 3.06 × 1021atoms of Au (s)
1 g of Na (s) = 
mol of Na (s)
atoms of Na (s)
= 0.262 × 1023 atoms of Na (s)
= 26.2 × 1021 atoms of Na (s)
1 g of Li (s) 
mol of Li (s)
atoms of Li (s)
= 0.86 × 1023 atoms of Li (s)
= 86.0 × 1021 atoms of Li (s)
1 g of Cl2 (g) 
mol of Cl2 (g)
(Molar mass of Cl2 molecule = 35.5 × 2 = 71 g mol–1)
molecules of Cl2 (g)
= 0.0848 × 1023 molecules of Cl2 (g)
= 8.48 × 1021 molecules of Cl2 (g)
As one molecule of Cl2 contains two atoms of Cl.
Number of atoms of Cl = 2× 8.48 × 1021 =16.96 × 1021 atoms of Cl
Hence, 1 g of Li (s) will have the largest number of atoms.
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