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A 12.0 g sample of gas occupies 19.2 L at STP. what is the of moles and molecular weight of this gas?​

User Jonno
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1 Answer

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At STP, 1 mole of an ideal gas occupies a volume of about 22.4 L. So if n is the number of moles of this gas, then

n / (19.2 L) = (1 mole) / (22.4 L) ==> n = (19.2 L•mole) / (22.4 L) ≈ 0.857 mol

If the sample has a mass of 12.0 g, then its molecular weight is

(12.0 g) / n14.0 g/mol

User Hcknl
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