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The density of gold is 19g/cm^3. If 1.9 ...

The density of gold is 19g/`cm^3`. If `1.9 xx 10^(-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)`

Text Solution

Verified by Experts

The correct Answer is:
D
Volume of gold used =`(m)/(d)=(1.9xx10^(-4))/(19)=1xx10^(-5) cm^(3)`
`therefore` Total no of particles =`(1xx10^(-5))/(4.19 xx10^(-18))=2.4 xx10^(12)`
Thus `2.4 xx10^(12)` particles of gold are persent in 1 litre or `10^(-3) m^(3)`
`therefore` Number of particles per `mm^(3) =(2.4 xx10^(12))/(10^(-3)xx10^(9))=2.4 xx10^(6)`
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