Question

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

Answer 1

`"33.6 L"`.

Explanation:

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"`

Answer 2

33.6 litres

Explanation:

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.