Question

how many atoms of silver are present in a silver wire weighing 5.4 grams​

Answer 1

Answer:

1 mole of Ag = 6.022 x 1023 atoms = 108 g

so, Atoms present in 5.4 g of Ag wire = 5.4 x 6.022 x 1023 / 108 = 3.01 x 1022 atoms

Answer 2

Explanation:

Mole = Given weight/Atomic Weight

= 5.4/107

= 0.0504 moles

No.of atoms = Na× n

= 6.022×10^23 × 0.0504

= 0.303 × 10^23 atoms