The density of solid argon is 1.65 g per cc at - 233° C. If the argon atom is assumed to be a sphere of radius 1.54 × 10-8 cm, what per cent of solid argon is apparently empty space? (Ar = 40)

Question

The density of solid argon is 1.65 g per cc at - 233° C. If the argon atom is assumed to be a sphere of radius 1.54 × 10-8 cm, what per cent of solid argon is apparently empty space? (Ar = 40) (A) 16.5% (B) 38% (C) 50% (D) 62%

Tardigrade Admin · Accepted Answer

Volume of one molecule =(4/3) π r3 =(4/3) π(1.54 × 10-8)3 cm 3 = 1.53 × 10-23 cm 3 Volume of all molecules in 1.65 g Ar =(1.65/40) × N0 × 1.53 × 10-23 =0.380 cm 3 Volume of solid containing 1.65 g Ar =1 cm 3 ∴ Empty space =1-0.380 =0.620 ∴ Per cent of empty space =62 %

Q. The density of solid argon is 1.65gpercc at - 233∘C. If the argon atom is assumed to be a sphere of radius 1.54×10−8cm, what per cent of solid argon is apparently empty space? (Ar=40)

16.5%

38%

50%

62%

Volume of one molecule =34​πr3 =34​π(1.54×10−8)3cm3 =1.53×10−23cm3 Volume of all molecules in 1.65g Ar =401.65​×N0​×1.53×10−23 =0.380cm3 Volume of solid containing 1.65gAr=1cm3 ∴ Empty space =1−0.380 =0.620 ∴ Per cent of empty space =62%

Q. The density of solid argon is $1.65 g p er cc$ at - $23 3^{\circ} C$ . If the argon atom is assumed to be a sphere of radius $1.54 \times 1 0^{- 8} c m$ , what per cent of solid argon is apparently empty space? $(A r = 40)$