Question

The no of silver atoms present in a 90% pure silver wire

Answer 1

Answer:

If 90% silver: (10 g) * 0.9 = 9 grams Ag

The atomic weight of Ag is 107.8682.

9 g Ag * (1 mol Ag/107.8682 g Ag) = 0.083435155 mols Ag

Then use the Avogadro constant to convert mol to number of atoms.

0.083435155 mols Ag * (6.022E23/1 mol) =

5.0246E23 atoms of Ag or 50 sextillion 246 quintillion atoms of silver.