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

Jun 12, 2018

$\text{33.6 L}$.

#### Explanation:

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

$p V = n R T$,
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 $\text{273.15 K}$ and pressure is $\text{1 atm}$.

We also already know the gas constant. In our case, we'll use the value of $\text{0.08206 L atm/K mol}$ 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:

$n = \text{number of moles" = "mass of sample"/"molar mass}$
$= \text{6.00 g"/"4.00 g/mol" = "1.50 mol}$

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

$p V = n R T$

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

$\text{= 33.6 L}$

Jun 12, 2018

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.