A 0.010 cm thick coating of copper is deposited on a sheet of 0.8 m2 t

1.	A 0.010 cm thick coating of copper is deposited on a sheet of 0.8 m2 total area. Calculate the number of copper atoms deposited on the sheet. (Density of copper = 7.2 g cm–3 , atomic mass of copper 63.5).
Answer» Area of sheet = 0.8 m² = 0.8 × 10⁴ cm² Thickness of coating 0.010 cm Volume of copper deposited = Area × thickness = 0.8 × 10⁴ × 0.010 = 80 cm³ Mass of copper deposited = Volume × density = 80 × 7.2 = 576 g \(\because\) 63.5 g of copper contain = 6.022 × 10²³atoms \(\therefore\) 576 g of copper will contain = \(\frac{6.022\times10^{23}}{63.5}\) x 576 = 54.62 × 10²³ = 5.462 × 10²⁴ atoms

A 0.010 cm thick coating of copper is deposited on a sheet of 0.8 m2 total area. Calculate the number of copper atoms deposited on the sheet. (Density of copper = 7.2 g cm–3 , atomic mass of copper 63.5).

Answer»

Area of sheet = 0.8 m² = 0.8 × 10⁴ cm²

Thickness of coating 0.010 cm

Volume of copper deposited = Area × thickness

= 0.8 × 10⁴ × 0.010

= 80 cm³

Mass of copper deposited = Volume × density

= 80 × 7.2

= 576 g

\(\because\) 63.5 g of copper contain = 6.022 × 10²³atoms

\(\therefore\) 576 g of copper will contain = \(\frac{6.022\times10^{23}}{63.5}\) x 576

= 54.62 × 10²³

= 5.462 × 10²⁴ atoms