The mass of 3.01 x 1023 molecules of gas is 18.0 grams. What volume does 12.0 grams of the gas occupy at STP

Answer 1

1 mol of a gas is 6.02E23 molecules so moles you have is
(3.01E23/6.02E23) = 0.5 mol and that has a mass of 18 g so 1 mol has a mass of 36 g. Then 36 g will have a volume of 22.4 L @ STP
22.4 L x 12/36 = volume occupied by 12 g