Mole concept (Avogadro application): How many water molecules are present in a drop of water of volume 0.0018 mL at room temperature? Use density ≈ 1 g/mL and molar mass of water = 18 g/mol for the calculation. Choose the correct value.

Given data

• Volume = 0.0018 mL; ρ(H₂O) ≈ 1 g/mL; M(H₂O) = 18 g/mol; N_A = 6.023 × 10²³ mol⁻¹.

Step-by-Step calculation
Mass = 0.0018 mL × 1 g/mL = 0.0018 gMoles = 0.0018 g ÷ 18 g/mol = 1.0 × 10⁻⁴ molNumber of molecules = 1.0 × 10⁻⁴ × 6.023 × 10²³ = 6.023 × 10¹⁹

Common pitfalls
Mixing up milliliters and liters or forgetting that density is ~1 g/mL for water at room temperature.

Final Answer
6.023 × 10^19