Question

What volume is occupied by 26.6 g of argon gas at a pressure of 1.29 atm and a temperature of 355 K ?

Answer 1

Answer:

V = 15 L

Explanation:

PV = nRT

P = 1.29 atm, T= 355 K mass argon 40g/mol

n = mass/ molecular mass; n = 26.6 g/40 g/mol; n = 0.665 mol

V = (0,665 mol x 0.082 x 355 K) / 1.29 atm = 15 L