1.

The density of mercury is `13.6 g mL^(-1)`. Calculate the approximate diameter of an atom of mercury assuming that each atom is occupying a cube of edge length equal to the diameter of the mercury atom.

Answer» Let the diameter of one mercury atom be `a`.
Volume of mercury molecule `=a^(3)` (volume of cube)
Volume of `N` mercury molecules `=a^(3)xxN`
Density `=(Mass)/(Volume)=(200)/(a^(3)xxN)=13.6`
`a^(3)=(200)/(13.6xxN)=(200)/(13.6xx6.023xx10^(23))`
or `a=3sqrt((200)/(13.6xx6.023xx10^(3)))`
`=2.9004xx10^(-8)cm`


Discussion

No Comment Found

Related InterviewSolutions