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^{-9}x10^{2}
cm
since 1 m = 10^{2} cm

= 5x10^{-7} cm

Volume = V = (4/3) x π x r^{3}

=
(4/3) x π x (5x10^{-7})^{3} cm^{3}

=
5.236x10^{-19} cm^{3}

Density of silver , d = 10.49 g/cm^{3}

So mass of silver is , m = density x volume

= 10.49 g/cm^{3} x 5.236x10^{-19}
cm^{3}

= 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 6.023x10^{23} (atoms/mol)

= 30.6x10^{3} atoms

Therefore the number of Ag nanoparticle having a 5 nm radius is
30.6x10^{3} atoms

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