Question

What volume does 3.15 g of He occupy at STP?

Answer 1

17.64L since I mole occupies 22.4L
And I mole of He weighs 4g
So 4g weigh 22.4L
And 3.15g weigh 22.4/4 x 3.15
17.64

Answer 2

22.4 L.

Step-by-step explanation:

At STP, one mole of any gas occupies 22.4 L.

First, we need to find how many moles of He are present in 3.15 g:

moles of He = mass of He / molar mass of He

The molar mass of He is 4.003 g/mol.

moles of He = 3.15 g / 4.003 g/mol = 0.787 mol

Now we can use the relation between moles and volume at STP:

0.787 mol He = 22.4 L He

1 mol He = 22.4 L / 0.787 mol He = 28.5 L/mol

Therefore, 3.15 g of He occupies:

volume of He = (moles of He) x (volume per mole of He)

volume of He = 0.787 mol x 28.5 L/mol = 22.4 L (approximately)

So 3.15 g of He occupies about 22.4 L at STP.