Question

if 125 mg of ar(g) is added to a 505mL sample of Ar(g) at stp, what volume will the sample occupy when the conditions of STP are restored?

Answer 1

The sampy will occupy a volume of 575mL

Step-by-step explanation:

Mass of argon added = 125mg = 125/1000 = 0.125g, initial volume = 505mL, MW of argon = 40g/mole

Number of moles of argon added = mass of argon added ÷ MW = 0.125 ÷ 40 = 0.003125mole

At STP, 1 mole of a gas contains 22.4L (22.4×1000mL = 22400mL) of the gas

Therefore, 0.003125mole of argon contains 0.003125 × 22400mL = 70mL

Volume added = 70mL

Volume the sample will occupy = initial volume + volume added = 505mL + 70mL = 575mL