Answer:
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