Question

what volume of 5.6 gram of Nitrogen gas occupies at STP​

Answer 1

Explanation:

given mass = 5.6g

molecular mass of n2 = 14 ×2 = 28g

stp = 22.4 L

volume = given mass / molecular mass × 22.4

= 5.6/28×22.4

= 4.48L

volume of 5.6 grams of n2 at stp = 4.48 L

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Answer 2

Answer: 5.6 gram of Nitrogen gas occupies 4.48 Litre at STP​

Explanation:

Molar mass of nitrogen gas (N₂) = 28 gm

Mass of nitrogen gas taken = 5.6 gm

Moles of nitrogen gas = Mass taken ÷ Molar mass

Moles of nitrogen gas = 5.6/28

Moles of nitrogen gas  =  0.2

At STP,

Volume occupied by 1 mole of gas = 22.4 litre

Volume occupied by 0.2 moles of gas = 22.4 × 0.2 litre

Volume occupied = 4.48 litre

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