The density of solid argon is 1.6 gm/ml at -233°C. If the atomic volume of argon is assumed to be 53×10−28, then what % of solid argon is apparently empty space? [Express your answer by dividing with 15]

Volume of solid argon =391.6=24.375/moleVolume of 1 mole atoms of argon =53×10−23×6.023×1023=10.038% vacant =24.375−10.03824.375×100=60% Ans :6015=4

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The density of solid argon is 1.6 gm/ml at $- 233 ° C$ . If the atomic volume of argon is assumed to be $\frac{5}{3} \times 10^{- 28}$ , then what % of solid argon is apparently empty space? [Express your answer by dividing with 15]

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answer is 4.

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Volume of solid argon $= \frac{39}{1.6} = 24.375 / mole$

Volume of 1 mole atoms of argon

$= \frac{5}{3} \times 10^{- 23} \times 6.023 \times 10^{23} = 10.038$

$\begin{array}{l} % vacant = \frac{24.375 - 10.038}{24.375} \times 100 \\ = 60 % Ans : \frac{60}{15} = 4 \end{array}$

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