A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
louis413
 2 years ago
During an diffusion experiment, it took 75 seconds for a certain number of moles of an unknown gas
to pass through a tiny hole. Under the same conditions, the same number of moles of oxygen gas
passed through the hole in 30 seconds. What is the molar mass of the unknown gas?
louis413
 2 years ago
During an diffusion experiment, it took 75 seconds for a certain number of moles of an unknown gas to pass through a tiny hole. Under the same conditions, the same number of moles of oxygen gas passed through the hole in 30 seconds. What is the molar mass of the unknown gas?

This Question is Closed

JFraser
 2 years ago
Best ResponseYou've already chosen the best response.0In general, the heavier a gas molecule is, the slower it moves, and the longer it takes to effuse (escape through a tiny hole in a barrier). Graham's Law describes the relationship between the molar mass of a gas and its relative rate of effusion.\[\frac {R_1}{R_2} = \sqrt{\frac{MM_2}{MM_1}}\] You know that gas 1 passed through the hole in 75 seconds, and that gas 2 took only 30 seconds. You also know that gas 2 is oxygen gas \O_2. Oxygen gas has a molar mass of 32g/mol. The only thing missing is MM1. Plug and solve
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.