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Hello! I would really appreciate any help on this chemistry question on effusion and diffusion. Thank you! :)

If a mixture composed of 3.2 grams of helium, 4.9 g oxygen, 5.5 g argon exerts a total pressure of 980 Torr, what is the pressure the helium exerts in the container in atm?

Respuesta :

rsd05

Answer:

709 torr

Explanation:

  1. Find the number of mols for each atom present. This gives us 0.76 mol Helium, 0.15 mol Oxygen (I assumed Oxygen was O2 for the context of this problem, since Oxygen gas is O2), and 0.14 mol Argon
  2. Find the total number of mols, which is 1.05
  3. Find the mol fraction for Helium. Take the mols of helium, 0.76, and place it over 1.05, giving us 0.76 / 1.05
  4. Multiply 0.76 / 1.05 to by 980 torr get around 709 torr