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)`

Answer

62

Explanation

Volume of one atom of `Ar = (4)/(3) pi r^(3)`
Also, number of atoms in 1.65g or one `mL = (1.65)/(40) xx 6.023 xx 10^(23)`
`:.` total volume of all atoms of Ar in solid state
`= (4)/(3) pi r^(3) xx (1.65)/(40) xx 6.023 xx 10^(23) = (4)/(3) xx (22)/(7) xx (1.54 xx 10^(-8))^(3) xx (1.65)/(40) xx 6.023 xx 10^(23) = 0.380 cm^(3)`
Volume of solid argon `= 1cm^(3)`
`:.` % empty space `= ([1 - 0.80])/(1) xx 100 = 62%`