The density of gold is `19 g //cm^(2)`.Find the volume of 95 g of gold.

Options

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

Answer

D

Explanation

Volume of gold dispersed in 1 L water`=(Mass)/(Density)= (1.9xx10^(-4)g)/(19"gm cm"^(-3))= 1xx10^(-5) cm^(3)`
Radius of gold sol particle
`= 10 nm=10 xx 10^(-7)cm = 10^(-6) cm`
Volume of gol sol particler`=(4)/(3)xx(22)/(7)xx(10^(-6)0^(3)=4.19xx10^(-18)cm^(18)`
:.No. of gold sol particles `=(1xx10^(5))/(4.19xx10^(-18))=2.38xx10^(12)`
:. No. of gold sol particles in one `mm^(3)`
`=(2.38xx10^(12))/(10^(6))=2.38xx10^(6)`