Ncert Exercises & Solutions

13.14 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].

The volume of each atom

= \frac{4}{3} π r^{3}

The volume of N number of molecules=

\frac{4}{3} π r^{3}

N … (i)

The volume of one mole of a substance =

\frac{M}{ρ}

… (ii)

\Rightarrow \frac{4}{3} {πr}^{3} N = \frac{M}{ρ}

\Rightarrow r = \sqrt[3]{\frac{3 M}{4 πρN}}

For carbon: M = 12.01 ×

10^{- 3}

kg,

ρ = 2.22 ×

10^{3}

m^{- 3}

r = {(\frac{3 \times 12.01 \times 10^{- 3}}{4 π \times 2.22 \times 10^{3} \times 6.023 \times 10^{23}})}^{\frac{1}{3}} = 1.29 Å

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

For gold: M = 197.00 ×

10^{- 3}

ρ = 19.32 ×

10^{3}

m^{- 3}

r = {(\frac{3 \times 197 \times 10^{- 3}}{4 π \times 19.32 \times 10^{3} \times 6.023 \times 10^{23}})}^{\frac{1}{3}} = 1.59 Å

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

For liquid nitrogen:

M = 14.01 ×

10^{- 3}

ρ = 1.00 ×

10^{3}

m^{- 3}

r = {(\frac{3 \times 14.01 \times 10^{- 3}}{4 π \times 1.00 \times 10^{3} \times 6.23 \times 10^{23}})}^{\frac{1}{3}} = 1.77 Å

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

For lithium:

M = 6.94 ×

10^{- 3}

ρ = 0.53 ×

10^{3} {kgm}^{- 3}

r = {(\frac{3 \times 6.94 \times 10^{- 3}}{4 π \times 0.53 \times 10^{3} \times 6.23 \times 10^{23}})}^{\frac{1}{3}} = 1.73 Å

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

For liquid fluorine:

M = 19.00 ×

10^{- 3}

ρ = 1.14 ×

10^{3} kg m^{- 3}

r = {(\frac{3 \times 19 \times 10^{- 3}}{4 π \times 1.14 \times 10^{3} \times 6.023 \times 10^{23}})}^{\frac{1}{3}} = 1.88 Å

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

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