Question

Here's A Question!!
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Calculate the number of Oxygen atoms present in 44.8 L of ozone gas at S.T.P.
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Answer 1

Answer:

MOLE CONCEPT

Explanation:

No. of atoms of oxygen in ozone gas ( O3 ) / Avagadro no. = volume of gas at stp / 22.4 L

no. of atoms = 44.8 / 22.4 × 6.023 × 10²³

No. of atoms = 2 × 6.023 × 10^23

Answer :- 12. 05 × 10²³ atoms in 44.8 L of Ozone gas

Answer 2

Answer:

$moles = \frac{mass}{molecularweight} = \frac{no}{na} \\ moles (stp) = \frac{volume}{22.4} = \frac{44.8}{22.4} = 2 \\ \\ 2 = \frac{no \: of \: molecules}{6.02 \times {10}^{23} } \\ no \: of \: molecues = 12.04 \times {10}^{23} \\ no \: of \: atomes \: = no \: of \: molecules \times atomicity \\ no \: of \: atomes = 12.04 \times {10}^{23} \times 3 = 36.12 \times {10}^{23}$

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