If 1.00 mole of a gas has a mass of 18 grams, and you have 15.0 grams of the gas, what would its volume be at STP?

Expert Answers

An illustration of the letter 'A' in a speech bubbles

Molar mass of the gas  = 18 g per mol

The mass of gas available = 15 g

Therefore, the number of moles present is given by,


Number of moles = mass / molar mass

Number of moles = 15 g/ 18 g per mol

                         = 5/6 mol

                         = 0.83333 mol.

STP - Standard Temperature & Pressure

This is 0 C and 1 bar according to IUPAC definition. At these conditions, the volume of 1 mol of ideal gas is 22.71 L/mol.

If we assume that the given gas is an ideal, volume of 1 mol at STP is 22.71 L/mol. But according to Avogadro Law volume, V is,

V = kn   Where k is constant and n is the number of moles.

If the volume of 15 g of gas is V, then,

22.71 = k x 1

V = k x (5/6)  (since it has only 5/6 moles)

V = 22.71 x (5/6) = 18.925 L

 

Therefore, volume of 15 g of the gas is 18.925 L.

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial Team