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 x 10²³ atoms of K    [Avogadro's law]

Therefore, 117 g of K

$= \dfrac{6.023 \times 10^{23}}{39} \times 117 \\[0.5em] = \dfrac{117}{39} \times 6.023 \times 10^{23} \\[0.5em] = 3 \times 6.023 \times 10^{23} \\[0.5em]$

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