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

ConceptSpecific Heat Capacities - Gases

#### 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) 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

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