The density of solid argon ( Ar =40 g / mol ) is 1.68 g / mL at 40 K. If the argon atom is assumed to be a sphere of radius =1.50 × 10-8 cm, what % of solid Ar is apparently empty space?

Question

The density of solid argon ( Ar =40 g / mol ) is 1.68 g / mL at 40 K. If the argon atom is assumed to be a sphere of radius =1.50 × 10-8 cm, what % of solid Ar is apparently empty space? (A) 35.64 (B) 64.36 (C) 74 % (D) None of these

Tardigrade Admin · Accepted Answer

Volume of all atoms in 1.68 g Argon =(1.68/40) × N A × (4/3) × π ×(1.5 × 10-8)3=0.3564 Volume of solid argon =1 cm 3 % Empty space =(1-0.3564) × 100=64.36

Q. The density of solid argon $(A r = 40 g / m o l)$ is $1.68 g / m L$ at $40 K$ . If the argon atom is assumed to be a sphere of radius $= 1.50 \times 1 0^{- 8} c m$ , what $%$ of solid $A r$ is apparently empty space?

$35.64$

$64.36$

$74%$

None of these

Volume of all atoms in $1.68 g$

Argon $= \frac{1.68}{40} \times N_{A} \times \frac{4}{3} \times π \times (1.5 \times 1 0^{- 8})^{3} = 0.3564$

Volume of solid argon $= 1 c m^{3}$

$%$ Empty space $= (1 - 0.3564) \times 100 = 64.36$

Q. The density of solid argon (Ar=40g/mol) is 1.68g/mL at 40K. If the argon atom is assumed to be a sphere of radius =1.50×10−8cm, what % of solid Ar is apparently empty space?

35.64

64.36

74%