1.

The density of gold is `19 g//cm^(3)`. If `1.9xx10^(4)` 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 beA. `1.9xx10^(12)`B. `6.3xx10^(14)`C. `6.3xx10^(10)`D. `2.4xx10^(6)`

Answer» Correct Answer - D
Volume of the gold dispresed in one litre water
`=("mass")/("density") = (1.9xx10^(-4)g)/(19g cm^(-3)) = 1 xx 10^(5)cm^(-3)`
Radius of gold sol particle
`=10nm = 10 xx 10^(-9)m = 10xx10^(-7) cm = 10^(-6)cm`
Volume of gold sol particle = `4/3 pi r^(3)`
`4/3 xx 22/7 xx (10^(-6))^(3) = 4.19xx10^(-18) cm^(3)`
No. of gold sol particle in `1 xx 10^(-5)cm^(3)`
No, of gold sol particle in `1 xx 10^(-5) cm^(3)`
`=(1xx10^(-5))/(4.19xx10^(-18)) = 2.38xx10^(12)`
No. of gold sol particle in one `mm^(3)`
`=(2.38xx10^(12))/(10^6) = 2.38 xx 10^(6)`.


Discussion

No Comment Found

Related InterviewSolutions