The diameter of an oxygen molecule is 2.94 ti | vedclass

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(D) The van der Waals constant '$b$' represents the excluded volume per mole of gas.It is given by the formula: $b = 4 	imes N_A 	imes V_{molecule}$

Vedclass · Accepted Answer

$32 	imes 10^{-6}$

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(D) The van der Waals constant '$b$' represents the excluded volume per mole of gas.It is given by the formula: $b = 4 	imes N_A 	imes V_{molecule}$,where $N_A$ is Avogadro's number and $V_{molecule} = \frac{4}{3} \pi r^3$.Given diameter $d = 2.94 	imes 10^{-10} \ m$,so radius $r = \frac{d}{2} = 1.47 	imes 10^{-10} \ m$.$b = 4 	imes (6.022 	imes 10^{23}) 	imes \frac{4}{3} 	imes 3.14 	imes (1.47 	imes 10^{-10})^3$.$b = 4 	imes (6.022 	imes 10^{23}) 	imes \frac{4}{3} 	imes 3.14 	imes (3.1765 	imes 10^{-30})$.$b \approx 32 	imes 10^{-6} \ m^3/mol$.

The diameter of an oxygen molecule is $2.94 \times 10^{-10} \ m$. The van der Waals gas constant '$b$' in $m^3/mol$ is:

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