Question

what is the volume occupied by 20 g of oxygen(in litre) at stp​

Answer 1

Answer:

Hi there,

◇ Key

1 mole of gas occupies 22.4L volume at STP

◇ Given

20 gm O2

20g ×( 1mol / 32gm ) × (22.4 L / 1mol ) = 11.2 L

◇ Other way

Number of mols = mass ÷ molar mass

= 20 ÷ 32 = 0.6 mol

1mol = 22L

0.6 mol = 11.2L

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Answer 2

Molecular mass of O2 = 32g
32g O2 occupies at STP occupies volume = 22.4L
Therefore,20g of O2 at STP will occupy volume = (22.4)/32 . (20)
= 14L