A 12.0 g sample of gas occupies 19.2 L at STP. what is the of moles and molecular weight of this gas?
1 answer:
At STP, 1 mole of an ideal gas occupies a volume of about 22.4 L. So if <em>n</em> is the number of moles of this gas, then
<em>n</em> / (19.2 L) = (1 mole) / (22.4 L) ==> <em>n</em> = (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) / <em>n</em> ≈ 14.0 g/mol
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