Answer
The volume occupied by 50.0 g of Ar gas at STP is equal to 28.0 L.
Work Step by Step
1. Setup the conversion factors:
For a gas at STP conditions, we can use the equality: $1 \space mole = 22.4 \space L$:
$\frac{1 \space mole \space (Ar)}{22.4 \space L \space (Ar)}$ and $\frac{22.4 \space L \space (Ar)}{1 \space mole \space (Ar)}$
Molar mass of $Ar$: 39.95 g/mole
$ \frac{1 \space mole \space (Ar)}{ 39.95 \space g \space (Ar)}$ and $ \frac{ 39.95 \space g \space (Ar)}{1 \space mole \space (Ar)}$
2. Calculate the volume of that sample:
$50.0 \space g \space (Ar) \times \frac{1 \space mole \space (Ar)}{ 39.95 \space g \space (Ar)} \times \frac{22.4 \space L \space (Ar)}{1 \space mole \space (Ar)} = 28.0 \space L \space (Ar)$
