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InterviewSolution
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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 ]. |
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Answer» Atomic mass of a substance = M Density of the substance = `rho` Avogadro’s number = `N = 6.023 × 10^23` Volume of each atom `=4/3pir^3` Volume of N number of molecules `=4/3pir^3` N … (i) Volume of one mole of a substance `= (M)/(rho)` `=4/3pir^3 N=(M)/(P)` ` therefore r=root3((3M)/(4pirhoN)` For carbon: `M = 12.01 × 10^(–3) kg,` `rho = 2.22 × 10^3 kg m^(–3)` `thereforer=((3xx12.01xx10^-3)/(4pixx2.22xx10^3xx6.023xx10^23))^(1/3)` = 1.29 Å Hence, the radius of a carbon atom is 1.29 Å. For gold: ` M=197.00XX10^(-3) kg` `rho=19.32xx10^(3) kg m^(-3)` `thereforer=((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=14.01xx10^(-3) kg ` `rho=1.00xx10^3 kg m^(-3)` `thereforer=((3xx14.01xx10^-3)/(4pixx1.00xx10^3xx6.023xx10^23))^(1/3)` =1.77 Å Hence, the radius of a liquid nitrogen atom is 1.77 Å. For lithium: `M=6.94 xx 10^(-3) kg ` `rho = 0.53xx10^3 kg m^(-3)` `thereforer=((3xx6.94xx10^-3)/(4pixx0.53xx10^3xx6.023xx10^23))^(1/3)` For liquid flurorine `M = 19.00 xx 10^(–3)` kg `rho = 1.14 × 103 kg m^(–3)` `thereforer=((3xx19xx10^-3)/(4pixx1.14xx10^3xx6.023xx10^23))^(1/3)` = 1.88 ÅHence, the radius of a liquid fluorine atom is 1.88 Å |
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