Number of atoms present in 224 , dm^3 of oxyg | vedclass

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(D) At $STP$,$1 \, mole$ of any gas occupies $22.4 \, L$ (or $22.4 \, dm^3$).Number of moles of $O_2 = \frac{224 \, dm^3}{22.4 \, dm^3/mol} = 10 \, mo

Vedclass · Accepted Answer

$1.2 	imes 10^{25}$

Vedclass · Answer

(D) At $STP$,$1 \, mole$ of any gas occupies $22.4 \, L$ (or $22.4 \, dm^3$).Number of moles of $O_2 = \frac{224 \, dm^3}{22.4 \, dm^3/mol} = 10 \, mol$.Since $O_2$ is a diatomic molecule,$1 \, mole$ of $O_2$ contains $2 \, moles$ of oxygen atoms.Total moles of oxygen atoms $= 10 	imes 2 = 20 \, mol$.Number of atoms $= 20 	imes 6.022 	imes 10^{23} \approx 1.2 	imes 10^{25}$ atoms.

Number of atoms present in $224 \, dm^3$ of oxygen gas at $STP$ is

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