# What mass of helium is needed to inflate a baloon to a volume of 5.50 L at STP ?

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1 mol He at STP = 22.4L

5.5L at STP = 5.5/22.4 = 0.246 mol He

Molar mass He = 4.00g/mol

0.246 mol = 0.246*4 = 0.982g He required

Quote answer as 1.0g He required.

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• 1 mole of ideal gas occupies 22.4 L. You have 5.5L.

So 5.5 L * 22.4 mol/L =123.2 mol of He.

Molar mass of He is 4 grams/mole ---> 123.2 mol *4g/mol = 492.8 grams

EDIT :Well lets see,

1 gram of helium

= 0.25 mol of He which will give off a volume of V = 0.25*R*298K/1atm = 6.14625 L

I forgot to compute the right amount:

(1 atm)(5.5 L)/ (R)(298) = 0.223714 mols = 0.894855 grams. There that is the right answer. And also, accounting for the unit fugacity of the gas.

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• At STP that is not going to be much of an inflation. But at STP, one mole will occupy 22.4 liters.

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• I was the one who MISTAKENLY put a thumbs up on the 500-gram answer.

Fat fingers! - sorry

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