180ml of a hydrocarbon diffuses through a porous membrane in 15 minutes while 120ml of SO₂ under similar conditions diffuses in 20 minutes. What is the molecular mass of hydrocarbon? and how
 

Dear student,

Please find below the solution to the asked query.

According to Graham's law of diffusion:

$\frac{r_1}{r_2}=\frac{\sqrt{M_2}}{\sqrt{M_1}}$

where

r = rate of diffusion
M = Molecular mass

$r = \frac{Volume effused}{time}=\frac{V}{t}$

So,

$$\begin{gathered} \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \\ \Rightarrow \frac{V_1/t_1}{V_2/t_2} = \sqrt{\frac{M_2}{M_1}} \\ \Rightarrow \frac{180/15}{120/20}=\sqrt{\frac{64}{M_1}} (Molecular mass of S{O}_2 = 64 g/mol) \\ \Rightarrow \frac{180\times 20}{120\times 15} = \frac{8}{\sqrt{M_1}} \\ \Rightarrow \sqrt{M_1} = \frac{8\times 120\times 15}{180\times 20} \\ \Rightarrow \sqrt{M_1} = 4 \\ On squaring both sides \\ \Rightarrow M_1 = 16 \end{gathered}$$

So, molecular mass of the hydrocarbon = 16 g/mol

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