Question

What volume will 20.0g of Argon occupy at STP?

Answer 1

Molar mass Argon = 39.948 g/mol

1 mol ------ 39.948 g
  mol ----- 20.0 g 

mol = 20.0 * 1 / 39.948

= 0.5006 moles

1 mol --------------------- 22.4 L ( at STP )
0.5006 moles ------------- L 

L = 0.5006 * 22.4

= 11.21 L

hope this helps!