Question

Question 1.28 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)

Class XI Some Basic Concepts of Chemistry Page 24

Answer 1

(i) 1 g Au = 1/197 mol = 1/197×6.022×1023 atoms


(ii) 1 g Na = 1/23 mol = 1/23×6.022×1023 atoms


(iii) 1 g Li = 1/7 mol = 1/7×6.022×1023 atoms


(iv) 1 g Cl2 = 1/71 mol = 1/71×6.022×1023 atoms


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

Answer 2

(i) 1g of Au ,
atomic mass of Au = 197 g/mol
so, no of mole of 1 g of Au = 1/197
so, no of atoms of 1g of Au = 1/197 × Avogadro's constant ( Let A ) = A/197

(ii) 1g of Na,
atomic mass of Na = 23g/mol
so, no of mole of 1g of Na = 1/23
so,no of atoms of 1g of Na = 1/23 × A = A/23

(iii) 1g of Li
atomic mass of Li = 7g/mol
so,no of mole of 1g of Li = 1/7
so, no of atoms of 1g of Li = A/7

(iv) 1g of Cl2(g)
molar mass of Cl2 = 71g/mol
so, no of mole of 1 g of Cl2 = 1/71
so, no of atoms of 1g of Cl2 = A/71

here, we see maximum number of atoms is A/7
so, 1g of Li has largest number of atoms