Calculate the no. of atoms of potassium present in 117 g. of K. [K = 39]

Gram atomic mass of K = 39

At s.t.p.,

39 g of K = 6.023 × 10²³ atoms of K [Avogadro’s law]

⇒ 117 g of K = 6.023 × 10^{23}\over 39×117

= 117 × 10^{23}\over 39 × 6.023 × 10²³

= 3 × 6.023 × 10²³

Hence, number of atoms in 117 g of K = 3 × 6.023 x 10²³