A sample of Xe takes 75 seconds to effuse out of a container. An unknown gas takes 37 seconds to effuse out of the identical container under identical conditions. What is the most likely identity of the uknown gas
the answer is O2. How do you solve this?
Oh, I see we have a case of the mysterious gas here! Well, let's put on our detective hats and figure this out, shall we?
First, let's establish some facts. We know that the sample of Xenon (Xe) took 75 seconds to effuse out of the container. And we also know that the unknown gas took only 37 seconds to effuse out of the identical container under identical conditions.
The rate of effusion is inversely proportional to the square root of the molar mass of the gas. In simpler terms, a lighter gas will effuse faster than a heavier gas.
So, let's compare the times it took for effusion: Xe took 75 seconds and the unknown gas took 37 seconds. Since the unknown gas is effusing faster, it must have a lower molar mass than Xe.
Ah, but which gas could it be? Well, looking at the periodic table, the only gas with a lower molar mass than Xenon (Xe) is Oxygen (O2).
So, my dear Watson, based on this information, the most likely identity of the unknown gas is indeed Oxygen (O2). Case closed!
To solve this problem, we can use Graham's Law of effusion. According to Graham's Law, the rate of effusion for two gases is inversely proportional to the square root of their molar masses.
Let's denote the molar mass of Xe as M(Xe) and the molar mass of the unknown gas as M(unknown).
Based on the given information, we have the following ratios for the rates of effusion:
Rate(Xe) / Rate(unknown) = √(M(unknown) / M(Xe))
We can set up the following equation using the given information:
75 seconds / 37 seconds = √(M(unknown) / M(Xe))
To solve for M(unknown), let's square both sides of the equation:
(75/37)^2 = M(unknown) / M(Xe)
Simplifying the left side:
2.49 = M(unknown) / M(Xe)
Now we can rearrange the equation to solve for M(unknown):
M(unknown) = 2.49 * M(Xe)
Since Xe is the heaviest noble gas with a molar mass of 131.29 g/mol, we can substitute this value into the equation:
M(unknown) = 2.49 * 131.29 g/mol = 327.15 g/mol (rounded to two decimal places)
We can compare this molar mass to the molar masses of known gases. The molar mass of oxygen (O2) is approximately 32 g/mol.
Since the calculated molar mass (327.15 g/mol) is significantly larger than the molar mass of oxygen (32 g/mol), we can conclude that the most likely identity of the unknown gas is O2.
To solve this problem, we can use Graham's Law of Effusion. According to Graham's Law, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
In this case, we have two gases: Xe and an unknown gas. Let's represent the molar mass of Xe as M(Xe) and the molar mass of the unknown gas as M(unknown).
Now, let's calculate the ratio of their effusion rates:
Rate(Xe) / Rate(unknown) = √[M(unknown) / M(Xe)]
Given that the effusion time for Xe is 75 seconds and the effusion time for the unknown gas is 37 seconds, we can write:
Rate(Xe) / Rate(unknown) = 75 / 37
Now we can solve for the ratio of their molar masses:
√[M(unknown) / M(Xe)] = 75 / 37
Squaring both sides of the equation, we get:
M(unknown) / M(Xe) = (75 / 37)^2
Now, we can substitute the values and calculate:
M(unknown) / M(Xe) ≈ 3.984
Since we want to know the identity of the unknown gas, we need to find the gas with a molar mass ratio close to 3.984. Looking at the periodic table, we find that the molar mass ratio of oxygen (O2) to xenon (Xe) is approximately 4.003.
Therefore, the unknown gas is most likely oxygen (O2).
Grahams law look that up in your text.
Xe has molweight of 131
sqrt(xxx/131)=37/75
xxx molecular weight= 131 (37/75)^2
amazing, it is 32 the same as O2.