The volume of a drop of water is 0.09 mL. Assuming the density of water to be 1.0 g mL ⁻¹ , the total number of electrons present in this drop of water is : (Given : N_A = 6.022 10²³ mol ⁻¹ )

Step-by-step solution

Mass of the water drop = Volume Density = 0.09 mL 1.0 g mL ⁻¹ = 0.09 g . Molar mass of water ( H ₂ O ) = 18 g mol ⁻¹ . Number of moles of water = 0.09 18 = 0.005 mol . One molecule of water ( H ₂ O ) contains 2 electrons from the hydrogen atoms and 8 electrons from the oxygen atom, giving a total of 10 electrons. Number of moles of electrons = 10 0.005 = 0.05 mol . Total number of electrons = 0.05 N_A = 0.05 6.022 10²³ = 3.011 10²² . Answer: 3.011 10²²