0.05 cm thick coating of silver is deposited on a plate of 0.05 m² area. The number of silver atoms deposited on plate are ________ × 10²³. (At mass Ag=108, d=7.9 g cm⁻³)

Solution

Chemistry ma’am · 3 steps

Get the volume of silver first, and convert the area to cm² before you do.

1. Match the units. The density is in g cm⁻³, so work in centimetres. Since 1 m² = 10⁴ cm², the area is 0.05 × 10⁴ = 500 cm².
2. Volume and mass. Volume of the coating = 500 × 0.05 = 25 cm³. Mass = 25 × 7.9 = 197.5 g.

   N = 197.5/108 × 6.022 × 10²³ = 1.829 × 6.022 × 10²³

3. So. N = 11.01 × 10²³ silver atoms, which rounds to 11.

The common slip: 0.11. Converting 0.05 m² with a factor of 100 instead of 10⁴ gives an area of 5 cm² and an answer of 0.11, not 11.

The idea to keep. For a coating, mass is area times thickness times density. Put every length in centimetres first, because m² to cm² is a factor of 10⁴, not 100.

Answer from the NTA final answer key. Solution by PrepWiser.