An unknown gas is found to diffuse through a porous membrane
2.9...

Question

An unknown gas is found to diffuse through a porous membrane 
2.92 times more slowly than H2. What is the molecular 
weight of the gas? 
(a) 17.0 g/mol (b) 5.84 g/mol 
(c) 8.52 g/mol

Accepted Answer

Welcome back everyone. An unidentified gas diffuses 1.33 times more slowly than chlorine gas through a porous membrane. What is the gasses molecular weight. So what we want to recall is that for our fusion of a gas, we would say that the rate of that gas is inversely proportional to its square root of the molar mass. So this is our infusion formula of a gas recall that a fusion is where gas particles are traveling through a small hole, which is where we have or small opening, which is why our prompt states we have a porous membrane. So what we want to do is write out a relationship where the rate of our unknown gas Divided by our rate of our chlorine gas is equal to the square root of our molar mass of our chlorine gas divided by the molar mass of our unknown gas. Where according to our prompt, this entire quotient should equal 1.33. So we are going to refer to our periodic tables to find the molar mass of chlorine gas. And we would see that according to our periodic table chlorine gas has a molar mass of 70.90 g per mole. So, plugging in our known into our equation will have that our square root. Sorry, So our square root Where we have our molar mass of chlorine gas being 70.90 g per mole Divided by our molar mass of our unknown gas is equal to 1.33. And to simplify this, we're going to first focus on taking the square root of both sides to get rid of that square root term. Where we will now be able to say that we have 70. g per mole divided by the molar mass of our unknown gas equal to a value of 1.7689. And then simplifying further, we're just going to recall that in algebra when we have diagonals, we can just trade places. So what we're going to have is that 70.90 g per mole divided by 1.7689 is equal to a value of 40 g per mole. And this would be our final answer as the molecular weight of our unknown gas. I hope everything I explained was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.

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