Test Index

Kinetic Theory

© examsnet.com
Question : 14 of 14
Marks: +1, -0
Given below are the densities of some solids and liquids. Give rough estimates of the sizes of their atoms :
 Substance  Atomic Mass (u)   Density (103 kg m−310^3\,\mathrm{kg}\,\mathrm{m}^{-3})
 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 rr be the radius of an atom. Then,
Volume of an atom, V=43πr3V = \frac{4}{3}\pi r^{3}
Volume of atoms in one mole of the substance, V=NV=N×43πr3V = N V = N \times \frac{4}{3}\pi r^{3}
where NN = Avogadro’s number If MM is atomic mass and rr, the density of the substance, then
N×43πr3=MρN \times \frac{4}{3}\pi r^{3} = \frac{M}{\rho} or r=(3M4πρN)1/3r = \left( \frac{3M}{4\pi\rho N} \right)^{1/3}
For carbon :M=12.01×10−3 kg: M = 12.01 \times 10^{-3}\,\mathrm{kg} ;
ρ=2.22×103 kg m−3\rho = 2.22 \times 10^{3}\,\mathrm{kg}\,\mathrm{m}^{-3}
∴r=(3×12.01×10−34π×2.22×103×6.023×1023)1/3\therefore r = \left( \frac{3 \times 12.01 \times 10^{-3}}{4\pi \times 2.22 \times 10^{3} \times 6.023 \times 10^{23}} \right)^{1/3}
=1.29×10−10 m=1.29 A˚= 1.29 \times 10^{-10}\,\mathrm{m} = 1.29\,\text{\AA}
For gold :M=197×10−3 kg: M = 197 \times 10^{-3}\,\mathrm{kg} ;
ρ=19.32×103 kg m−3\rho = 19.32 \times 10^{3}\,\mathrm{kg}\,\mathrm{m}^{-3}
∴r=(3×197×10−34π×19.32×103×6.023×1023)1/3\therefore r = \left( \frac{3 \times 197 \times 10^{-3}}{4\pi \times 19.32 \times 10^{3} \times 6.023 \times 10^{23}} \right)^{1/3}
=1.59×10−10 m=1.59 A˚= 1.59 \times 10^{-10}\,\mathrm{m} = 1.59\,\text{\AA}
For nitrogen :M=14.01×10−3 kg: M = 14.01 \times 10^{-3}\,\mathrm{kg};
ρ=1.0×103 kg m−3\rho = 1.0 \times 10^{3}\,\mathrm{kg}\,\mathrm{m}^{-3}
∴r=(3×14.01×10−34π×1.0×103×6.023×1023t)1/3\therefore r = \left( \frac{3 \times 14.01 \times 10^{-3}}{4\pi \times 1.0 \times 10^{3} \times 6.023 \times 10^{23}} t \right)^{1/3}
=1.77×10−10 m=1.77 A˚= 1.77 \times 10^{-10}\,\mathrm{m} = 1.77\,\text{Å}
For lithium :M=6.94×10−3 kg: M = 6.94 \times 10^{-3}\,\mathrm{kg} ;
ρ=0.53×103 kg m−3\rho = 0.53 \times 10^{3}\,\mathrm{kg}\,\mathrm{m}^{-3}
∴r=(3×6.94×10−34π×0.53×103×6.023×1023)1/3\therefore r = \left( \frac{3 \times 6.94 \times 10^{-3}}{4\pi \times 0.53 \times 10^{3} \times 6.023 \times 10^{23}} \right)^{1/3}
=1.73×10−10 m=1.73 A˚= 1.73 \times 10^{-10}\,\mathrm{m} = 1.73\,\text{\AA}
For fluorine :M=19×10−3 kg: M = 19 \times 10^{-3}\,\mathrm{kg};
ρ=1.14×103 kg m−3\rho = 1.14 \times 10^{3}\,\mathrm{kg}\,\mathrm{m}^{-3}
∴r=(3×19×10−34π×1.14×103×6.023×1023)1/3\therefore r = \left( \frac{3 \times 19 \times 10^{-3}}{4\pi \times 1.14 \times 10^{3} \times 6.023 \times 10^{23}} \right)^{1/3}
=1.88×10−10 m=1.88 A˚= 1.88 \times 10^{-10}\,\mathrm{m} = 1.88\,\text{\AA}
© examsnet.com
Go to Question:
Prev Question