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 - what - of solid Ar is apparently empty space-A- 64-36B- 74C- 35-64D- none of these

The correct option is A nbsp;64.36Volume of all atoms in 1.68 g of Ar nbsp; = 1.6840 x6 x1023 nbsp;x4/3 x22/7 x(1.503 x10 24) nbsp; = 0.3564 Vol. of s ...

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Standard XII

Chemistry

Density

The density o...

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 Å , what % of solid Ar is apparently empty space?

35.64

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64.36

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none of these

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Solution

The correct option is B 64.36
Volume of all atoms in 1.68 g of Ar =
$\frac{1.68}{40}$ x6 x $10^{23}$ x4/3 x22/7 x( $1 . 50^{3}$ x $10^{- 24}$ ) = 0.3564
Vol. of solid Ar = 1 ${cm}^{3}$
% Vol. of empty space
= (1 - 0.3564) x100 = 64.36

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Similar questions

Q. 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 Å , what % of solid Ar is apparently empty space?

Q. The density of solid argon is 1.4 g/mL at

- 233^{\circ} C

If the argon atom is assumed to be sphere of radius

2.0 \times 10^{- 8}

cm, What percentage of solid argon is apparently empty space? (atomic wt. Of Ar = 40)

(take

N_{A} = 6 \times 10^{23}

Q. 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 x

10^{- 8}

cm, what % of solid Ar is apparently empty space? (use

N_{A} = 6 \times 10^{23}

)

Q. The density of solid argon is

1.65 g

per cc at

- 233

. If the argon atom is assumed to be a sphere of radius

1.54 \times 10^{- 8} c m

, what percent of solid argon is apparently empty space?

(A r = 40)

Q. The density of solid argon is

1.65

g mL

^{- 1}

- 233^{\circ} C

. If the argon atom is assumed to be the sphere of radius

1.54 \times 10^{- 8}

cm, what percentage of solid argon is apparently empty space?

[Atomic mass of Ar is

40

amu.]

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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 Å , what % of solid Ar is apparently empty space?

The correct option is B 64.36Volume of all atoms in 1.68 g of Ar = 1.6840 x6 x1023 x4/3 x22/7 x(1.503 x10−24) = 0.3564 Vol. of solid Ar = 1 cm3 % Vol. of empty space = (1 - 0.3564) x100 = 64.36

The correct option is B 64.36
Volume of all atoms in 1.68 g of Ar =
$\frac{1.68}{40}$ x6 x $10^{23}$ x4/3 x22/7 x( $1 . 50^{3}$ x $10^{- 24}$ ) = 0.3564
Vol. of solid Ar = 1 ${cm}^{3}$
% Vol. of empty space
= (1 - 0.3564) x100 = 64.36