What mass of helium is needed to inflate a baloon to a volume of 5.50 L at STP ?
Solve please :)?
- Trevor HLv 78 years agoFavorite Answer
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.
- 8 years ago
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
so no your answers are not right either.
Correction to my answer,
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.
- cat loverLv 78 years ago
At STP that is not going to be much of an inflation. But at STP, one mole will occupy 22.4 liters.
- az_lenderLv 78 years ago
I was the one who MISTAKENLY put a thumbs up on the 500-gram answer.
Fat fingers! - sorry
The right answer is
4 g times (5.5 L/22.4 L) = ONE gram.