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A gaseous mixture contains oxygen and another unknown gas in the molar ratio of 4:1 diffuses through a porous plug into 245 seconds, under similar conditions same volume of oxygen takes 220 seconds to diffuse. What is the molecular mass of the unknown gas?

I used the Graham's law of diffusion to find the molecular mass of mixture. i.e. $$\frac{r_\text{mix}}{r_{\ce{O2}}}=\frac{220}{245}=\sqrt{\frac{M_{\ce{O2}}}{M_\text{mix}}}$$ $$M_\text{mix}=39.6$$

But what's next? How do I calculate the $M_\text{gas}$?

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    $\begingroup$ You know the molar ratio of oxygen to the unknown gas is 4:1, so you can write an equation for the molar mass of the unknown gas in terms of $M_\text{mix}$. $\endgroup$ – a-cyclohexane-molecule Aug 3 '18 at 14:14
  • $\begingroup$ How do I do that? $\endgroup$ – The ReBel Aug 3 '18 at 17:02
  • $\begingroup$ Think about what $M_\text{mix}$ is. If I have 0.8 moles of $A$ and 0.2 moles of $B$, what is the (molar) mass of the mixture in terms of the molar masses $m_A$ and $m_B$ of $A$ and $B$ respectively? $\endgroup$ – a-cyclohexane-molecule Aug 3 '18 at 17:08
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After getting relative molecular mass ie $39.6$,

$$\mathrm{M_{net} = \frac{(n_1\times M_1 + n_2\times M_2)}{(n_1 + n_2)}} \tag{i}$$

Given that the molar ratio of gases is $4:1$, the molar ratio of oxygen is $\mathrm{4x}$ and that of the unknown gas is $\mathrm x$. Plug $\mathrm{n_1 = 4x,\ n_2 = x,\ M_1= 32,\ M_2 = M,\ M_{net} = 39.6}$ in equation (i) and get $\mathrm M$.

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