The density of gold is 19`g//cm^(3)`. If 1.8`xx10^(-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:A. `1.9xx10^(12)`B. `6.3xx10^(14)`C. `6.3xx10^(10)`D. `2.4xx10^(6)`

The density of gold is 19`g//cm^(3)`. If 1.8`xx10^(-4)`g of gold is di

1.	The density of gold is 19`g//cm^(3)`. If 1.8`xx10^(-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:A. `1.9xx10^(12)`B. `6.3xx10^(14)`C. `6.3xx10^(10)`D. `2.4xx10^(6)`
Answer» Correct Answer - D 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)`.