The density of solid argon is `1.65g//mL ` at `-233^(@)C` . If the argon atom is assumed to be sphere of radius `1.54xx10^(-8)cm`, what percentage of solid argon is apparentaly empty space ? `(At. Wt. of Ar=40)`

Explanation

density of ` 1.65 " g mL" ^(-1)` means that 1.65 g of solid argon has volume =1 mL
Volume of one atom of Ar =` 4/3 pi r^(3)`
No. Of atoms in 1.65 g or 1 mL of solid `Ar = ((1.65)/40 "mol") (6.023 xx10^(23) "mol"^(-1)) = ( 1.65xx6.023xx10^(23))/40`
Total volume of all atoms Ar in 1 mL of Ar. `4/3 pi r^(3)xx(1.65xx6.023xx10^(23))/40`
=` 4/3xx22/7 xx(1.54 xx10^(-8))^(3) xx (1.65xx6.023xx10^(23))/40 `
0.380 ` cm^(3) or` 0.380 mL.
Empty space = (1 -0.380) mL= 0.620 mL
% empty spae =` ( 0.620)/ 1 xx 100 = 62 %`