0 . 05cm thick coating of silver is deposited on a plate of 0 . 05...

Volume of silver coating $= 0 . 05 \times 0 . 05 \times 10000 = 25{cm}^{3}$

Given, Density =$7 . 9g{cm}^{- 3}$

Density = Mass/ Volume 

Mass of silver deposited $= 25 \times 7 . 9g$

Moles of silver atoms = mass / molar mass $= \dfrac{25 \times 7 . 9}{108}$

Number of silver atoms $= \dfrac{25 \times 7 . 9}{108} \times 6 . 023 \times {10}^{23}$

$= 11 . 01 \times {10}^{23}$

Ans. $11$