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 =`4/3pir^3`
No. of atom in 1.65 g or 1 mL of solid Ar=`(1.65/40"mol")(6.023xx10^23 "mol"^(-1))=(1.65xx6.023xx10^23)/40`
`therefore` Total volume of all atoms Ar in 1 mL of Ar =`4/3pir^3 xx (1.65xx6.023xx10^23)/40`
`=4/3xx22/7xx(1.54xx10^(-8))^3 xx (1.65xx6.023xx10^23)/40`
`=0.380 cm^3` or 0.380 mL
Volume of solid Ar = 1 mL
`therefore` Empty space =(1-0.380) mL =0.620 mL
`therefore` % empty space =`0.620/1xx100` =62%