Question

Calculate the number of Ag nanoparticle having a 5 nm radius.?

Answer 1

Given radius of Ag, r = 5 nm = 5x10^-9 m = 5x10^-9x10² cm            since 1 m = 10² cm

                                         = 5x10^-7 cm

Volume = V = (4/3) x π x r³

           = (4/3) x π x (5x10^-7)³ cm³

           = 5.236x10^-19 cm³

Density of silver , d = 10.49 g/cm³

So mass of silver is , m = density x volume

                                   = 10.49 g/cm³ x 5.236x10^-19 cm³

                                   = 5.49x10^-18 g                           

No of moles,n = Mass / Atomic weight

                    = 5.49x10^-18 g / 107.9 (g/mol)

                    = 5.09x10^-20 mol

No of atoms = number of moles x Avogadro's number

                   = 5.09x10^-20 mol x 6.023x10²³ (atoms/mol)

                   = 30.6x10³ atoms

Therefore the number of Ag nanoparticle having a 5 nm radius is 30.6x10³ atoms