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Question 28

Which one of the following will have 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)

Answer

(i) Gram atomic mass of Au= 197 g 

Or

197g of Au contains = 6.022 x 1023

 

Therefore 1gm of Au contains = 6.022 x 1023/197*1 = 3.06 x 1021 atoms

 

(ii) Gram atomic mass of Na = 23 g

Or

23 g of Na contains atoms = 6.022x 1023

Or

1gm of Na contains atoms = 6.022x1023 /23 *1 = 26.2 x1021 atoms

 

(iii) Gram atomic mass of Li = 7

Or

7g of Li contains atoms = 6.022 x 1023

Or

1g of Li contains atoms = 6.022 x 1023/7 *1= 86.0 x 1021 atoms

 

(iv) Gram atomic mass of Cl = 71 Or 71g of Cl contains atoms = 6.022x1023

Or

1 g of Cl contains atoms = 6.022x1023 /71 * 1= 8.48 x 1021 atoms

 

Hence, 1 g of Li (s) will have the largest number of atoms.

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