Question

# The density of solid argon is 1.4 g/mL at −233∘C If the argon atom is assumed to be sphere of radius 2.0×10−8 cm, What percentage of solid argon is apparently empty space? (atomic wt. Of Ar = 40) (take NA=6×1023

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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 = 2.0×10−8cm, we can say volume occupied by atoms =43πr3×1.440×6×1023=43×227×(2.0×10−8)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 =[1−0.704]1×100=29.6

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