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The critical density of the gas ${CO}_{2}$ is $0.44 g {cm}^{- 3}$ at a certain temperature. If r is the radius of the molecule, $r^{3}$ in ${cm}^{3}$ is approximately : (N is Avogadro number)

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$\frac{100}{π N}$

$\frac{25}{π N}$

$\frac{20}{π N}$

$\frac{6}{π N}$

answer is C.

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Detailed Solution

The value of $r^{3}$ in ${cm}^{3}$ is

$0.44 = \frac{44}{N} \times \frac{1}{3 \times 4 \times \frac{4}{3} π r^{3}} \Rightarrow r^{3} = \frac{6.25}{π N}$

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