The density of solid argon is 1.65 g/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 apparently empty space? (Ar = 40)

Options

A. 0.84
B. 0.78
C. 0.62
D. 0.48

Answer

C

Explanation

First determine the no. of atoms of Ar present in 1.65 g, from this calculate the total volume occupied by all atoms of Ar. We can determine % empty space by subtracting the volume of solid Ar taken and volume occupied by Ar.
Volume of one atom of `Ar=(4)/(3)pir^(3)`
Also, number of atoms in 1.65 g or one `cm^(3)=(1.65)/(40)xx6.02xx10^(23)=N`
`therefore `Total volume of al atoms of Ar in solid state=`(4)/(3)pir^(3)N=(4)/(3)xx3.14xx(1.54xx10^(-8))^(3)xxN=0.380cm^(3)`
As the volume of solid argon taken=`1cm^(3)` `therefore` % empty space`=([1-0.380])/(1)xx100=62%`.