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Question

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)

Gold

Nitrogen (liquid)

Lithium

Fluorine (liquid)

12.01

197.00

14.01

6.94

19.00

2.22

19.32

1.00

0.53

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].

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Solution

Substance

Radius (Å)

Carbon (diamond)

Gold

Nitrogen (liquid)

Lithium

Fluorine (liquid)

1.29

1.59

1.77

1.73

1.88

Atomic mass of a substance = M

Density of the substance = ρ

Avogadro’s number = N = 6.023 × 1023

Volume of each atom

Volume of N number of molecules N … (i)

Volume of one mole of a substance = … (ii)

N =

For carbon:

M = 12.01 × 10–3 kg,

ρ = 2.22 × 103 kg m–3

= 1.29 Å

Hence, the radius of a carbon atom is 1.29 Å.

For gold:

M = 197.00 × 10–3 kg

ρ = 19.32 × 103 kg m–3

= 1.59 Å

Hence, the radius of a gold atom is 1.59 Å.

For liquid nitrogen:

M = 14.01 × 10–3 kg

ρ = 1.00 × 103 kg m–3

= 1.77 Å

Hence, the radius of a liquid nitrogen atom is 1.77 Å.

For lithium:

M = 6.94 × 10–3 kg

ρ = 0.53 × 103 kg m–3

= 1.73 Å

Hence, the radius of a lithium atom is 1.73 Å.

For liquid fluorine:

M = 19.00 × 10–3 kg

ρ = 1.14 × 103 kg m–3

= 1.88 Å

Hence, the radius of a liquid fluorine atom is 1.88 Å.


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