The density of gold is 19g/`cm^(3)`. If `1.9times10^(-4)g` of gold is dispersed in one litre of water to give a sol having spherical gold particles of radius 10nm, then the number of gold particles per `mm^(3)` of the sol will be

Options

A. `1.9times10^(12)`
B. `6.3times10^(14)`
C. `6.3times10^(10)`
D. `2.4times10^(6)`

Answer

D

Explanation

Volume of gold present in solution `=("Mass of gold")/("Density of gold")=(1.9times10^(-4)g)/(19g*cm^(-1))=0.1times10^(-4)cm^(3)`.
For spherical particles of gold with radius equal to 10nm
The volume of each particles`=(4)/(3)pir^(3)=(4)/(3)times(22)/(7)times(10times10^(-7)cm)^(3)=(88)/(21)times10^(-18)cm^(3)`
number of gold particles `=("Volume of gold in solution")/("Volume of each particle")=(0.1times10^(-4)cm^(3))/((88)/(21)times10^(-18)cm^(3))`
`=(21)/(88)times10^(13)` particles`=2.4times10^(12)` particles.
`2.4times10^(12)` particles of gold are present in `1000cm^(3)`(litre)
`therefore` Number of particles present per `mm^(3)=(2.4times10^(24))/(10^(6))=2.4times10^(6)`