The density of gold is $$19\,g/c{m^3}.$$  If $$1.9 \times {10^{ - 4}}g$$   of gold is dispersed in one litre of water to give a sol having spherical gold particles of radius $$10\,nm,$$  then the number of gold particles per $$m{m^3}$$  of the sol will be :

Solution :
$$\eqalign{ & {\text{Volume of gold dispersed in }}1{\text{ }}L{\text{ water}} \cr & {\text{ = }}\frac{{mass}}{{Density}} = \frac{{1.9 \times {{10}^{ - 4}}g}}{{19gm\,c{m^{ - 3}}}} = 1 \times {10^{ - 5}}\,c{m^3} \cr & {\text{Radius of gold sol particle}} \cr & = 10\,nm = 10 \times {10^{ - 7}}\,cm = {10^{ - 6}}\,cm \cr & {\text{Volume of gol sol particle}} = \frac{4}{3}\pi {r^3} \cr & = \frac{4}{3} \times \frac{{22}}{7} \times {\left( {{{10}^{ - 6}}} \right)^3} = 4.19 \times {10^{ - 18}}c{m^3} \cr & \therefore \,\,{\text{No}}{\text{. of gold sol particles in}}\,\,1 \times {10^{ - 5}}c{m^3} \cr & = \frac{{1 \times {{10}^{ - 5}}}}{{4.19 \times {{10}^{ - 18}}}} = 2.38 \times {10^{12}} \cr & \therefore \,\,{\text{No}}{\text{. of gold sol particles in one}}\,m{m^3} \cr & = \frac{{2.38 \times {{10}^{12}}}}{{{{10}^6}}} \cr & = 2.38 \times {10^6} \cr} $$