Question

If 0.213 moles of argon occupies a volume of 652 mL at a particular temp and pressure, what volume would it occupy if 0.162 moles of argon were added at the same conditions?

Answer 1

If 0.213 moles of argon occupies 652 mL, adding 0.162 moles (for a total of 0.375 moles) would occupy approximately 1165 mL under the same conditions, using a direct proportion from Avogadro's Law.

Step-by-step explanation:

The question asks about the volume a certain amount of argon gas would occupy under constant temperature and pressure, which relates to the Chemistry principles involving the Ideal Gas Law (PV = nRT). Since the temperature and pressure are constant, we can relate the initial and final states of the argon gas using a direct proportion because V/n = constant (Avogadro's Law). Starting with 0.213 moles of argon occupying a volume of 652 mL, the addition of 0.162 moles of argon gives us a total of 0.213 + 0.162 = 0.375 moles of argon. Using the direct proportion, if 0.213 moles occupy 652 mL, then 0.375 moles would occupy an unknown volume (V).

To find V, you can set up the following ratio: (0.213 moles / 652 mL) = (0.375 moles / V mL). Solving for V, you get V = (0.375 moles * 652 mL) / 0.213 moles, which calculates to approximately 1165 mL (rounded to nearest whole number). Therefore, if you add 0.162 moles of argon to the original sample, the new volume it would occupy at the same conditions is 1165 mL.