Kinetic Theory

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Question : 14
Total: 14
Given below are the densities of some solids and liquids. Give rough estimates of the sizes of their atoms :
 Substance  Atomic Mass (u)   Density (103kgm3)
 Carbon (diamond)  12.01  2.22
 Gold  197.0  19.32
 Nitrogen (liquid)  14.01  1.00
 Lithium  6.94  0.53
 Fluorine (liquid)  19.00  1.14
Solution:  
Let r be the radius of an atom. Then,
Volume of an atom, V=
4
3
π
r3

Volume of atoms in one mole of the substance, V=NV=N×
4
3
π
r3

where N = Avogadro’s number If M is atomic mass and r, the density of the substance, then
N×
4
3
π
r3
=
M
ρ
or r=(
3M
4πρN
)
13

For carbon :M=12.01×103 kg ;
ρ=2.22×103kgm3
r=(
3×12.01×103
4π×2.22×103×6.023×1023
)
13
=1.29×1010m=1.29 Å
For gold :M=197×103kg ;
ρ=19.32×103kg m3
r= (
3×197×103
4π×19.32×103×6.023×1023
)
13
=1.59×1010m=1.59 Å
For nitrogen :M=14.01×103 kg;
ρ=1.0×103 kgm3
r= (
3×14.01×103
4π×1.0×103×6.023×1023
t
)
13
=1.77×1010m=1.77Å
For lithium :M=6.94×103 kg ;
ρ=0.53×103kg m3
r= (
3×6.94×103
4π×0.53×103×6.023×1023
)
13
=1.73×1010m=1.73 Å
For fluorine :M=19×103 kg;
ρ=1.14×103 kgm3
r=(
3×19×103
4π×1.14×103×6.023×1023
)
13
=1.88×1010 m=1.88Å
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