Question

How many Number of mole s and molecules are present in 48 gm O2 as well as calculate volume occupied by 48gm O2​

Answer 1

$Answer$

We know that PV=nRT,

Therefore V=nRTP

So we need to find number of moles of both the gasses.

No. of moles =wt.Mol.wt.

No. of moles of O2=1632=12

No. of moles of O3=48−1648=3248=23

Total No. of moles of gasses = 12+23=76

Therefore total volume =76×RTP=7×RT6×P=7×0.082×2736×1≈26.11L

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Or we can proceed in this way :

We know that total no. moles of gasses =76

Volume of 1 mol. of gas at STP =22.4L

Therefore volume of 76 mol. of gas at STP =76×22.4=26.13L

Therefore my final answer to this question is 26.1L ….……:-)

Answer 2

Answer:

48g of O2

Has 9.03 x 10^23 molecules

Is 1.5 moles of O2

Occupies 33.6 liters in terms volume

Explanation:

Molecular wt of O2 = 32

1 mole of O2 = 32g

? moles of O2 = 48g

48/32 = 1.5 moles of O2

One mole of O2 = 22.4l at STP

1.5 moles of O2 = ? liters

22.4 x 1.5 = 33.6l

One mole = 6.023 x 10^23 molecules

1.5 moles = ? molecules

6.023 x 1.5 x 10^23 = 9.03 x 10^23 molecules