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))/(19 gm cm^(-3)) = 1 xx 10^(-5) cm^(3)`.
Radius of gold sol particle = 10 nm = `10xx10^(-9) m = 10 xx 10^(-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))`
`=2.38xx10^(12)`
No. of gold sol particle in one `mm^(3)`
`=(2.38xx10^(22))/(10^(6)) = 2.38 xx 10^(6)`.