A gas of unknown molecular mass was allowed to effuse
through a s...

Question

A gas of unknown molecular mass was allowed to effuse
through a small opening under constant-pressure conditions.
It required 105 s for 1.0 L of the gas to effuse. Under
identical experimental conditions it required 31 s for 1.0 L
of O2 gas to effuse. Calculate the molar mass of the
unknown gas. (Remember that the faster the rate of effusion,
the shorter the time required for effusion of 1.0 L; in
other words, rate is the amount that diffuses over the time
it takes to diffuse.)

Accepted Answer

Hi everyone for this problem, we're told to consider the following experiment. A krypton gas was allowed to refuse through a small hole under constant pressure. The two liter sample of Krypton took 96 seconds to refuse another gas of unknown identity went through the same conditions. It took a two liter sample of the unknown gas 55.5 seconds to refuse. We need to calculate the molar mass of the unknown gas. So for this problem, we need to recall Graham's law of effusion and it tells us what the ratio is between the time that it takes a gas to infuse and its molar mass. And so that ratio is the time that it takes for gas to over the time that it takes for gas. One to refuse is directly proportional to the square root of the molar mass of Gas to over the molar mass of Gas one. And so let's define our variables here. T. One Is going to represent or one. Let's start there. We'll say one is going to represent krypton and two is going to represent are unknown. Okay, so based off of that information, we can say that T1 which represents Krypton. The time that it takes to refuse is 96 seconds and T two which represents our unknown, Takes 55.5 seconds to defuse. And then molar mass of one, which is Krypton. When we look that up, its molar mass is 83.8 g per mole. And our molar mass of our second gas is unknown. And that's what we're trying to solve for. So we'll write X there. So let's go ahead and plug in what we know based off of what was given. Okay. And so t. Two we said is 55.5 seconds. T one is 96 seconds and this is equal to the square root of molar mass to which is X Over Moller Mass one 83 point eight g per mole. So remember here we're solving for X and this X is going to tell us the molar mass of our unknown. So in order to get rid of the square root, we need to square both sides. Okay, so 55.5 divided by squared. So we're going to divide this out and square it. So I'll write it here. So this is going to square and this is going to square. And when we do that we get zero point 334 seconds is equal to X over 83.8 g per mole. So the squaring the right side just got rid of the square root. Okay, so now we can solve for X by multiplying both sides by 83. g per mole. And when we do that We get X is equal to 28 g per mole. This is our final answer. This is the molar mass of our unknown gas and that is the end of this problem. I hope this was helpful

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