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 ( |
---|---|---|
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 =
π r 3
Volume of atoms in one mole of the substance,V = N V = N ×
π r 3
whereN = Avogadro’s number If M is atomic mass and r , the density of the substance, then
N ×
π r 3 =
or r = (
) 1 ∕ 3
For carbon: M = 12.01 × 10 − 3 k g ;
ρ = 2.22 × 10 3 k g m − 3
∴ r = (
) 1 ∕ 3 = 1.29 × 10 − 10 m = 1.29 Å
For gold: M = 197 × 10 − 3 k g ;
ρ = 19.32 × 10 3 k g m − 3
∴ r = (
) 1 ∕ 3 = 1.59 × 10 − 10 m = 1.59 Å
For nitrogen: M = 14.01 × 10 − 3 k g ;
ρ = 1.0 × 10 3 k g m − 3
∴ r = (
t ) 1 ∕ 3 = 1.77 × 10 − 10 m = 1.77 Å
For lithium: M = 6.94 × 10 − 3 k g ;
ρ = 0.53 × 10 3 k g m − 3
∴ r = (
) 1 ∕ 3 = 1.73 × 10 − 10 m = 1.73 Å
For fluorine: M = 19 × 10 − 3 k g ;
ρ = 1.14 × 10 3 k g m − 3
∴ r = (
) 1 ∕ 3 = 1.88 × 10 − 10 m = 1.88 Å
Volume of an atom,
Volume of atoms in one mole of the substance,
where
For carbon
For gold
For nitrogen
For lithium
For fluorine
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