Question

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

Answer 1

Given: The densities and atomic mass of some solids and liquids are:

Substance	Atomic Mass ( u )	Density ( 10 3  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

The volume of each atom is given as,

V= 4 3 π r 3

Where, the radius of the atom is r.

The volume of all atoms in one mole of substance is given as,

V T = 4 3 π r 3 × N A (1)

Where, N A is the Avogadro’s number.

Also, the volume of one mole of substance is given as,

V T = M ρ (2)

Where, M is the atomic mass of the substance and ρis the density of the substance.

From equation (1) and (2), we get

M ρ = 4 3 π r 3 × N A r= ( 3M 4πρ N A ) 1 3 (3)

For carbon:

Substituting the given values in equation (3), we get

R= [ 3×12.01× 10 −3 4π×2.22× 10 3 ×6.023× 10 23 ] 1 3 = [ 2.14× 10 −30 ] 1 3 =1.29× 10 −10  m =1.29  Α o

Thus, the radius of the carbon atom is 1.29  A o .

For gold:

Substituting the given values in equation (3), we get

R= [ 3×197× 10 −3 4π×19.32× 10 3 ×6.023× 10 23 ] 1 3 =1.59× 10 −10  m =1.59  A o

Thus, the radius of the gold atom is 1.59  A o .

For nitrogen (liquid):

Substituting the given values in equation (3), we get

R= [ 3×14.01× 10 −3 4π×6.023× 10 23 ×1.00× 10 3 ] 1 3 =1.77× 10 −10  m =1.77  A o

Thus, the radius of the nitrogen (liquid) atom is 1.77  A o .

For lithium:

Substituting the given values in equation (3), we get

R= [ 3×6.94× 10 −3 4π×6.023× 10 23 ×0.53× 10 3 ] 1 3 =1.73× 10 −10  m =1.73  A o

Thus, the radius of the lithium atom is 1.73  A o .

For fluorine (liquid):

Substituting the given values in equation (3), we get

R= [ 3×19× 10 −3 4π×6.023× 10 23 ×1.14× 10 3 ] 1 3 =1.88× 10 −10  m =1.88  A o

Thus, the radius of the fluorine (liquid) atom is 1.88  A o .