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The density of mercury is 13.6 g/ml. Calculate approximately the 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.

Shane Macguire , 10 Years ago
Grade upto college level
anser 3 Answers
Deepak Patra

Last Activity: 10 Years ago

Hello Student,
Please find the answer to your question
Avogadro’s number = 6.023 * 1023
At. wt. of mercury(Hg) = 200
∵ In 1 g of Hg, the total number of atom
= 6.023 *1023/200 = 6.023 *1023/2 *102
= 3.0115 *1021 = 3.012 * 1021
∵ Density of Mercury (Hg) = 13.6 g/c.c.
∴ mass of 3.012 * 1021 atoms = 1/3.012 *1021
Now volume of 1 atom of mercury (Hg)
= 1/3.012 *1021 *13.6 c.c. = 103 *10/3012 *1021 * 136 c.c.
= 10-17/3012 *136 c.c. 10-17/409632 c.c. = 1000000 *10-23/409632 c.c.
= 2.44 *1023 c.c.
Since each mercury atom occupies a cube of edge length equal to its diameter, therefore,
Diameter of one Hg atom = (2.44 * 10-23)1/3 cm
= (24.4 * 10-24)1/3 cm.
= 2.905 * 10-8 cm ≡ 2.91 Å

Thanks
Deepak patra
askIITians Faculty

Naveen Raaz

Last Activity: 7 Years ago

Density of Hg=13.6g/lAtomic Mass of Hg=200amuV=Mass/densityLet the diameter be a, thena^3=(200×10^23)/13.6×6.023So,a=(24.4×10^-24)^-1/3Therefore, a=2.9×10^-8 cm or 2.9angstrom.

vandana Sharma

Last Activity: 5 Years ago

Suppose the length of the side of the cuve is X cm., i.e.,  the diameter of one Hg atom. 
•volume occupied by 1 Hg atom=X^3 cc
And Mass of one Hg atom =13.6×X^3g
•Mass of one Hg atom=at.Wt/Avconstant=200/6.023×10^23 g 
(•mass of 1 mole of atom is the atomic weight in g,  and 1 mole contain the Av.  Const.  Of atoms) 
Hence,  13.6×X^3=200/6.023×10^23
X^3=200/13.6×6.023×10^23=2.44×10^-23
X=2.9×10^-8 cm
 

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