CBSE (Science) Class 11CBSE
Share
Notifications

View all notifications

Given Below Are Densities of Some Solids and Liquids. Give Rough Estimates of the Size of Their Atoms: - CBSE (Science) Class 11 - Physics

Login
Create free account


      Forgot password?

Question

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

Substance Atomic Mass (u) Density (10Kg 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].

Solution 1

Substance Radius (Å)
Carbon (diamond) 1.29
Gold 1.59
Nitrogen (liquid) 1.77
Lithium 1.73
Fluorine (liquid) 1.88

Atomic mass of a substance = M

Density of the substance = ρ

Avogadro’s number = N = 6.023 × 1023

Volume of each atom = `4/3 pi r^3`

Volume of N number of molecules = `4/3 pir^3N` ... (i)

Volume of one mole of a substance = `M/rho` ... (ii)

`4/3 pir^3 N = M/rho`

`:. r = root(3)((3m)/(4pirhoN))`

For carbon

M = 12.01 × 10–3 kg,

ρ = 2.22 × 103 kg m–3

`:. r = ((3xx12.01xx10^(-3))/(4pixx2.22xx10^3xx6.023xx 10^23))^(1/3)` = 1.29 Å

For gold:

M = 197.00 × 10–3 kg

ρ = 19.32 × 103 kg m–3

:. r = `((3xx197xx10^3)/(4pixx19.32xx10^3xx6.023xx10^23))^(1/3)` = 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

`:. r = ((3xx14.01xx10^(-3))/(4pixx1.00xx10^3xx 6.23 xx10^23))^(1/3)` = 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

`:. r= ((3xx6.94xx10^(-3))/(4pixx0.53xx10^3xx6.23xx10^23))^(1/3)` = 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

`:. r= ((3xx19xx10^(-3))/(4pixx1.14xx10^3xx6.023xx10^23))^(1/3)` = = 1.88 Å

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

Solution 2

In one mole of a substance, there are 6.023 x 1023 atoms

`:. (4/3 pi R^3) xx 6.023 xx 10^23 = M/rho`

or ` R = [(3M)/(4pirho xx 6.023xx10^23)]^(1/3)`

For carbon  = `12.01 xx 10^(-3)` kg and `rho = 2.22 xx 10^3 "kg m"^(-3)`

`:. R = [(3xx12.01xx10^(-3))/(4xx3.14xx2.22xx10^3xx6.023xx10^23)]^(1/3)`

= `1.29 xx 10^(10) m` = 1.29 Å

For gold, M = `197 xx10^(-3)` kg and `rho = 19.32 xx 10^3 "kg m"^(-3)`

`:. R = [(3xx197xx10^(-3))/(4xx3.14xx19.32xx10^3 xx 6.023 xx 10^(23))]^(1/3)`

= 1.59 xx 10^(-10) m = 1.59 Å

For nitrogen (liquid), `M= 14.01 xx 10^(-3)` kg and `rho = 1.00 xx 10^3 "kg  m"^(-3)`

`R = [(3xx14.01xx10^(-3))/(4xx3.14xx100xx10^3xx6.023xx10^23)]^(1/3)`

 = `1.77 xx 10^(-10)m` = 1.77  Å

For lithium, M= `6.94 xx 10^(-3)` kg, `rho = 0.53 xx 10^3 "kg m"^(-3)`

`:. R = [(3xx6.94xx10^(-3))/(4xx3.14xx0.53xx10^3xx6.023xx10^(23))]^(1/3)`

`= 1.73 xx 10^(-10)`  = 1.73 Å

For fluorine (liquid), M = `19.00 xx 10^(-3)` kg, `rho = 1.14 xx 10^3 "kg m"^(-3)`

`:. R = [(3xx19.00xx10^(-3))/(4xx3.14xx1.14xx10^3xx6.023xx10^(23))]^(`1/3)`

`= 1.88 xx 10^(-10)m` = 1.88 Å

  Is there an error in this question or solution?

APPEARS IN

Video TutorialsVIEW ALL [1]

Solution Given Below Are Densities of Some Solids and Liquids. Give Rough Estimates of the Size of Their Atoms: Concept: Specific Heat Capacities - Gases.
S
View in app×