Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms :

 Substance Atomic Mass (u) Density (103 Kg m-3) Carbon (diamond) 12.01 2.22 Gold 197.00 19.32 Nitrogen (liquid) 14.01 1.00 Lithium 6.94 0.53 Fluorine (liquid) 19.00 1.14

[Hint: Assume the atoms to be ‘tightly packed’ in a solid or liquid phase, and use the known value of Avogadro’s number. You should, however, not take the actual numbers you obtain for various atomic sizes too literally. Because of the crudeness of the tight packing approximation, the results only indicate that atomic sizes are in the range of a few Å].

Asked by Pragya Singh | 1 year ago |  131

##### Solution :-

If r is the radius of the atom then the volume of each atom = ($$\dfrac{4}{3}$$)πr3

Volume of all the substance = ($$\dfrac{4}{3}$$)πrx N =$$\dfrac{M}{ρ}$$

M is the atomic mass of the substance

ρ is the density of the substance

One mole of the substance has 6.023 x 1023 atoms

r = $$( \dfrac{3M}{4πρ \times 6.023 \times 1023})^ \dfrac{1}{3}$$

For carbon, M = 12. 01 x 10-3 kg and ρ = 2.22 x 103 kg m-3

R = $$( \dfrac{3 \times 12. 01 \times 10^{-3}}{4 \times 3.14 \times 2.22 \times 10^3 \times 6.023 \times 10^{23}})^ \dfrac{1}{3}$$

$$( \dfrac{36.03 \times 10^{-3}}{167.94 \times 10^{26}})^ \dfrac{1}{3}$$

1.29 x 10 -10 m = 1.29 Å

For gold, M = 197 x 10-3 kg and ρ = 19. 32 x 103 kg m-3

R = $$( \dfrac{3 \times 197 \times 10^{-3}}{4 \times 3.14 \times 19.32 \times 10^3 \times 6.023 \times 10^{23}})^ \dfrac{1}{3}$$

= 1.59 x 10 -10 m = 1.59 Å

For lithium, M = 6.94 x 10-3 kg  and ρ = 0.53 x 103 kg/m3

R = $$( \dfrac{3 \times 6.94 \times 10^{-3}}{4 \times 3.14 \times 0.53 \times 10^3 \times 6.023 \times 10^{23}})^ \dfrac{1}{3}$$

= 1.73 x 10 -10 m = 1.73 Å

For nitrogen (liquid), M = 14.01 x 10-3 kg  and ρ = 1.00 x 103 kg/m3

R = $$( \dfrac{3 \times 14.01 \times 10^{-3}}{4 \times 3.14 \times 1.00 \times 10^3 \times 6.023 \times 10^{23}})^ \dfrac{1}{3}$$

= 1.77 x 10 -10 m = 1.77 Å

For fluorine (liquid), M = 19.00 x 10-3 kg  and ρ = 1.14 x 103 kg/m3

R = $$( \dfrac{3 \times 19 \times 10^{-3}}{4 \times 3.14 \times 1.14 \times 10^3 \times 6.023 \times 10^{23}})^ \dfrac{1}{3}$$

= 1.88 x 10 -10 m = 1.88 Å

Answered by Abhisek | 1 year ago

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