The density of gold is 19 g/cm³. If 1.9x10^-4 g of gold is dispersed in one litre of water to give a sol having spherical gold particles of radius l0 nm, then the number of gold particles per mm³ of the sol will be :

Volume of gold dispersed in 1 Lwater
$=\frac{\text{ Mass }}{\text{ Density }}=\frac{1.9\times {10}^{-4}\mathrm{g}}{19\mathrm{gm}{\mathrm{cm}}^{-3}}=1\times {10}^{-5}{\mathrm{cm}}^3$
Radius of gold sol particle
$=10\mathrm{nm}=10\times {10}^{-7}\mathrm{cm}={10}^{-6}\mathrm{cm}$
Volume of gol sol particle = $\frac{4}{3}{\mathrm{\pi r}}^3$
$=\frac{4}{3}\times \frac{22}{7}\times {\left({10}^{-6}\right)}^3=4.19\times {10}^{-18}{\mathrm{cm}}^3$
$\therefore$No. of gold sol particles in $1\times {10}^{-5}{\mathrm{cm}}^3$
$=\frac{1\times {10}^{-5}}{4.19\times {10}^{-18}}=2.38\times {10}^{12}$
$\therefore$No. of gold sol particles in one mm³
$=\frac{2.38\times {10}^{12}}{{10}^6}=2.38\times {10}^6$