What would be the volume of 6.00 g of helium gas at STP?

To answer this question, we'll need to use the Ideal Gas Law:

#pV = nRT#,
where #p# is pressure, #V# is volume, #n# is the number of moles #R# is the gas constant, and #T# is temperature in Kelvin.

The question already gives us the values for #p# and #T#, because helium is at STP. This means that temperature is #"273.15 K"# and pressure is #"1 atm"#.

We also already know the gas constant. In our case, we'll use the value of #"0.08206 L atm/K mol"# since these units fit the units of our given values the best.

We can find the value for #n# by dividing the mass of helium gas by its molar mass:

#n = "number of moles" = "mass of sample"/"molar mass"#
#= "6.00 g"/"4.00 g/mol" = "1.50 mol"#

Now, we can just plug all of these values in and solve for #V#:

#pV = nRT#

#V = (nRT)/p = ("1.50 mol" xx "0.08206 L atm/K mol" xx "273.15 K")/"1 atm"#

#"= 33.6 L"#

Assuming that helium behaves as an ideal gas, we know that a mole of ideal gas occupies 22.4 litres at STP.

The atomic weight of helium is 4 g/mol, so 6 g is 1.5 moles.

The volume occupied at STP, therefore, is 22.4 x 1.5 = 33.6 litres.