Nitrogen and an unknown gas is leaked into a common vessel... The gaseous mixture contains 0.4 moles of nitrogen. Molecular mass of the other gas is?
Calculate total moles using the ideal gas law and solving for $n$:

$$n = {4.15\;{\rm bar}\cdot 3\;{\rm L}\over {0.0831\;{\rm L}\,{\rm bar}\,{\rm K}^{-1}\,{\rm mol}^{-1}}\cdot 300\;{\rm K}} = 0.5\;{\rm mol}$$

Given 0.4 mol $\ce{N2}$, moles of unknown gas = 0.5 mol - 0.4 mol = 0.1 mol.

Rate of effusion of $\ce{N2} = r_{\ce{N2}} = 0.4\;{\rm mol}/10\;{\rm min} = 0.04\;{\rm mol}\cdot{\rm min}^{-1}$.

Rate of effusion of unknown ${\rm X} = r_{\rm X} = 0.1\;{\rm mol}/10\;{\rm min} = 0.01\;{\rm mol}\cdot{\rm min}^{-1}$.

Using Graham's law of diffusion:

$${r_{\ce{N2}}\over r_{\rm X}} = \sqrt{M_{\rm X}\over M_{\ce{N2}}}$$

we solve for $M_{\rm X}$:

$$M_{\rm X} = {M_{\ce{N2}}\times (r_{\ce{N2}})^{2}\over (r_{\rm X})^{2}}$$

and plugging in numbers we get the molecular mass of the unknown gas:

$$M_{\rm X} = {(28\,{\rm g}\cdot{\rm mol}^{-1})\times (0.04)^{2}\over (0.01)^{2}} = 448\,{{\rm g}\over{\rm mol}}$$

where the units on the effusion rates have been omitted for clarity and because they cancel.