During an effusion experiment, oxygen gas passed through a tiny hole 2.5 times faster than the same number of moles of another gas under the same conditions, what is the molar mass of the unknown gas?

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During an effusion experiment, oxygen gas passed through a tiny hole 2.5 times faster than the same number of moles of another gas under the same conditions, what is the molar mass of the unknown gas?

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Answer 1 · 2016-09-15T20:03:39

$M_{unknown} = 5.12 \frac{g}{mol}$ $M_"unknown" = 5.12 g / "mol"$

Explanation:

This problem can be solved using Graham's Law of Effusion. Graham discovered that the effussion rate of a gas is inversely proportional to the square root of its molar mass.

$\frac{r_{1}}{r_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}$ $r_1/r_2 = sqrt (M_2/M_1)$

Let's use this to solve the problem.

$r_{unknown} = 2.5 \times r_{oxygen}$ $r_"unknown" = 2.5 times r_"oxygen"$

$\frac{r_{unknown}}{r_{oxygen}} = \sqrt{\frac{M_{oxygen}}{M_{unknown}}}$ $(r_"unknown")/r_"oxygen" = sqrt (M_"oxygen"/M_"unknown")$

$\frac{r_{unknown}}{r_{oxygen}} = 2.5 = \sqrt{\frac{M_{oxygen}}{M_{unknown}}}$ $(r_"unknown")/r_"oxygen" = 2.5 = sqrt (M_"oxygen"/M_"unknown")$

$\frac{r_{unknown}}{r_{oxygen}} = {2.5}^{2} = (\frac{M_{oxygen}}{M_{unknown}})$ $(r_"unknown")/r_"oxygen" = 2.5^2 = (M_"oxygen"/M_"unknown")$

$M_{unknown} = (\frac{M_{oxygen}}{{2.5}^{2}})$ $M_"unknown" = (M_"oxygen"/2.5^2)$

$M_{unknown} = (\frac{16 \frac{g}{mol}}{{2.5}^{2}})$ $M_"unknown" = ((16 g / "mol" )/2.5^2)$

$M_{unknown} = 5.12 \frac{g}{mol}$ $M_"unknown" = 5.12 g / "mol"$

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During an effusion experiment, oxygen gas passed through a tiny hole 2.5 times faster than the same number of moles of another gas under the same conditions, what is the molar mass of the unknown gas?

1 Answer

Explanation:

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