# 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?

For 10 minutes each, from two identical holes, nitrogen and an unknown gas is leaked into a common vessel of 3L capacity at 300K. The resulting pressure is 4.15 bar. If the gaseous mixture contains 0.4 moles of nitrogen, then molecular mass of the unknown gas is?

attempt: According to Graham's law of effusion, $$\frac{r_1}{r_2}=\frac{V_1}{V_2}=\sqrt{\frac{M_2}{M_1}}$$ where $V_i$ is the volume of gas effused. Let $V_1$ and $M_1$ be volume and molecular mass of nitrogen. $$V_1=22.4\times0.4=8.96;M_1=28$$ Substituting, $${V_2}=8.96\sqrt{\frac{28}{M_2}}$$ and $V_1+V_2=3$, so $$8.96(1+\sqrt{\frac{28}{M_2}})=3$$ This is not possible since $\sqrt{x}\ge0$. What am I doing wrong?

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.

• This is the same answer given in my book. But am I making any conceptual mistakes? – Aditya Dev Jul 23 '15 at 19:18
• The thing that I see is you're incorrectly calculating volume (it can't be greater than 3 L, so multiplying 0.4 mol times 22.4 L/mol is not the place to start). I would say use what is given in the problem: they have explicitly given a time, a total pressure, a total volume, and the identity of one gas. From there, you should go to calculating moles of unknown. Conceptually, I would sum it up as look at what you're given and try to work the problem using those givens, and check your assumptions along the way (like the volume issue). – Todd Minehardt Jul 23 '15 at 19:24