Question

the number of silver atoms present in a 90% pure silver wire weighing 10 gram is​

Answer 1

Answer:

If 90% silver:(10g) *0.9 grams ag

The atomic weight of Ag is 107.8682.

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

Then use the Avagardo constant is convert mol to number of atoms.

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

5.0246E23 atoms lg Ag or so sexitillion 246 quintillion atoms of silver

Explanation:hope u get the right answer

Answer 2

Answer:

The correct answer is  5.0×10^22

Explanation:

Mass of pure silver=  90/100 x 10g = 9g

​ ∴ Moles of silver = (mass)/(molar mass)

                           = 9/108 = 1/12 (On simplifying we get 1/12)

Number of silver atoms= (moles of silver) x (Avagadro number) =1/12×6.022×10^23  

=0.5×10^23

=5×10^22

Therefore, the number of atoms present in 90% pure silver wire weighing 10 g is 5 x 10^22

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