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)

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Solution

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)

atoms of Cl₂ (g)

= 0.0848 × 10²³ atoms of Cl₂ (g)

= 8.48 × 10²¹ atoms of Cl₂ (g)

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

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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)

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)