# 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?

Sep 15, 2016

${M}_{\text{unknown" = 5.12 g / "mol}}$

#### 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.

${r}_{1} / {r}_{2} = \sqrt{{M}_{2} / {M}_{1}}$

Let's use this to solve the problem.

${r}_{\text{unknown" = 2.5 times r_"oxygen}}$

$\left({r}_{\text{unknown")/r_"oxygen" = sqrt (M_"oxygen"/M_"unknown}}\right)$

$\left({r}_{\text{unknown")/r_"oxygen" = 2.5 = sqrt (M_"oxygen"/M_"unknown}}\right)$

$\left({r}_{\text{unknown")/r_"oxygen" = 2.5^2 = (M_"oxygen"/M_"unknown}}\right)$

M_"unknown" = (M_"oxygen"/2.5^2)

M_"unknown" = ((16 g / "mol" )/2.5^2)

${M}_{\text{unknown" = 5.12 g / "mol}}$