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Question

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

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Answer

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

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