Question

Which one of the following will have largest number of atoms?

Options

• 1 g Au (s)

• 1 g Na (s)

• 1 g Li (s)

• 1 g of Cl₂(g)

Answer 1

1 g Li (s)

Explanation:

(i) 1 g of Au (s) `=1/197`mol of Au (s)

`=(6.022xx10^23)/197` atoms of Au (s)

= 3.06 × 10²¹atoms of Au (s)

(ii) 1 g of Na (s) `=1/23`mol of Na (s)

`=(6.022xx10^23)/23` atoms of Na (s)

= 0.262 × 10²³ atoms of Na (s)

= 26.2 × 10²¹ atoms of Na (s)

(iii) 1 g of Li (s) `=1/7` mol of Li (s)

`=(6.022xx10^23)/7` atoms of Li (s)

= 0.86 × 10²³ atoms of Li (s)

= 86.0 × 10²¹ atoms of Li (s)

(iv) 1 g of Cl₂ (g) `=1/71` mol of Cl₂ (g)

(Molar mass of Cl₂ molecule = 35.5 × 2 = 71 g mol^–1)

`=(6.022xx10^23)/71` 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.