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VITEEE 2013 Paper

Section: Chemistry

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Question : 59 of 120

Marks: +1, -0

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 x

10^{-8}

cm, what per cent of solid argon is apparently empty space? (Ar = 40)

16.5%
38%
50%
62%

Solution:

Volume of one molecule

=

\frac{4}{3} \pi r^3

=

\frac{4}{3}

π

(1.54 \times 10^{-8})^3 \text{cm}^3

= 1.53 ×

10^{-23} \text{cm}^3

Volume of molecules in 1.65 g Ar

=

\frac{1.65}{40}

×

N_0

× 1.53 ×

10^{-23}

= 0.380

\text{cm}^3

Volume of solid containing 1.65 gAr = 1

\text{cm}^3

∴ Empty space = 1-0.380 = 0.620

∴ Percent of empty space = 62%

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