Question

Chemistry mcq question 3? I have no idea how to go about it?

Answer 1

Atomic wt of Ag=107g/mol
Thus 0.216/107=0.002moles of Ag
Na=6.022×10^23
Thus 0.002×6.022×10^23
=1.205×10^21 atoms /mol
To find per cm divide by 150
Thus 1.205×10^21/150=8.033×10^18

Answer 2

Hi there!

Atomic weight of Silver = 107.8682 g/mol 

0.216g / 107.8682 g/mol = 0.002 moles

There are 6.02 × $10^{23}$ atoms/mole in a mole of anything (Avogadro number) 

0.002 mol × (6.02 × $10^{23}$ atoms/mole) = 1.205 × $10^{21}$ atoms of [Ag] in your Medal 

Divide by 150 cm³ :

= 8.033 × $10^{18}$ Silver atoms per cm²

Hence,
Option [ A ] - 8.033 × $10^{18}$ is Correct

[ Thank you! for asking the question. ]
Hope it helps!