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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. |
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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` |
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