The density of gold is `19g//cm^3`. If `1.9xx10^-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 `mm^3` of the sol will be:

Explanation

Volume of gold dispersed in 1 L of water `=("Mass")/("Density")=(1.9xx10^(-4))/(19 g cm^(-3))= 10^(-5)cm^(3)`
Radius of gold particle `=10 nm =10xx10^(-9)m=10^(-8)m=10^(-6)cm`
Volume of gold particle `=4/3pi r^(3)=4/3xx22/7 xx(10^(-6))^(3)=4.19 xx10^(-18)cm^(3)`
Thus, No. of gold particles in `10^(-5)cm^(3)=(10^(-5))/(4.19xx10^(-18))=2.38xx10^(12)`
These are the particles present in 1 L of the sol
`1 L =1000 cm^(3)=1000xx(10^(-2)m)^(3)=10^(-3) m^(3)`
`=10^(-3)xx(1000 mm0^(3) =10^(6) mm^(3)`
`therefore` No. of gold particles in `1 mm^(3)` of the sol `=(2.38xx10^(12))/(10^(6))=2.38xx10^(6)`