Question

What is the volume occupied by 24.2 g of argon gas at a pressure of 1.29 atm and a temperature of 374 K ?

Answer 1

The equation is PV=nRT. R is a constant which is equal to [0.0806 (L*atm)/(K*mol)]First, you have to make sure all of your givens are in the correct unit. Everything is correct except for the grams of Argon. The atomic mass of argon is 39.9g. To change Argon from grams to moles, (24.2g)*(1mol/39.9g). So the number of moles equals 0.60mol.

Now, using the equation V=(nRT)/P, plug in your numbers.
V=(0.60mol)(0.0806(L*atm/k*mol))(374K)/1.29... atm

All of the units should cancel out except for L.
The answer is 14.02L.