CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The density of solid argon is 1.4 g/mL at 233C If the argon atom is assumed to be sphere of radius 2.0×108 cm, What percentage of solid argon is apparently empty space? (atomic wt. Of Ar = 40)

(take NA=6×1023

Open in App
Solution

The number of Ar atoms in 40 g (1 mole) =6×1023
Number of atom in 1.4 g=1.440×6×1023
As density = 1.4 g/mL or 1.4 g/cm3
Number of atoms in 1 cm3 = No. of atoms in 1.4 g=1.440×6×1023 atoms.
Since the volume of each atom is 43πr3 where r = 2.0×108cm, we can say volume occupied by atoms
=43πr3×1.440×6×1023=43×227×(2.0×108)3×1.440×6×1023=0.704 cm3
per 1 cm3 of solid 0.704 cm3 volume is occupied by the atoms.
Hence % of empty space =[10.704]1×100=29.6

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Order of Magnitude
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon