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:

Options

A. `1.9xx10^(12)`
B. `6.3xx10^(14)`
C. `6.3xx10^(10)`
D. `2.4xx10^(6)`

Answer

D

Explanation

Volume of the gold dispersed in one litre water`=(mass)/(density)`
`=(1.9xx10(-4)gm)/(19" gm "cm^(-3))=1xx10^(-5)cm^(3)`
Radius of gold sol particle `=10nm=10xx10^(-9)m`
`=10xx10^(-7)cm=10^(-6)cm`
Volume of the gold sol particle `=(4)/(3)pir^(3)`
`=(4)/(3)xx(22)/(7)xx(10^(-6))^(3)=4.19xx10^(-18)cm^(3)`
No. of gold sol particle in `1xx10^(-5)cm^(3)=(1xx10^(-5))/(4.19xx10^(-18))`
No. of gold sol particle in one `nm^(3)`
`=(2.38xx10^(12))/(10^(6))=2.38xx10^(6)`.