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. (Atomic mass of Ag = 108, density = 7.9 g cm⁻³)

Silver Coating Numerical | Mole Concept | JEE Mains PYQ

Q. 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. (Atomic mass of Ag = 108, density = 7.9 g cm⁻³)

Step by Step Solution

Step 1: Convert area into cm²

1 m² = 10⁴ cm² Area = 0.05 × 10⁴ = 500 cm²

Step 2: Calculate volume of silver coating

Volume = Area × Thickness = 500 × 0.05 = 25 cm³

Step 3: Calculate mass of silver deposited

Mass = Density × Volume = 7.9 × 25 = 197.5 g

Step 4: Calculate moles of silver

Moles of Ag = 197.5 / 108 ≈ 1.83 mol

Step 5: Calculate number of atoms

Number of atoms = 1.83 × 6.02 × 10²³ ≈ 11 × 10²³

Final Answer: 11

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