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Concentration Terms - An Introduction (w/w etc)

180 mL of a h...

Question

$180 m L$ of a hydrocarbon diffuses through a porous membrane in $15$ minutes while $120 m L$ of $S O_{2}$ under identical conditions diffuses in $20$ minutes. What is the molecular mass of hydrocarbon?

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Solution

$r_{1} =$ rate of diffusion of hydrocarbon $= \frac{180}{15} m L m i n^{- 1}$
$r_{2} =$ rate of diffusion of $S O_{2} = \frac{120}{20} m L m i n^{- 1}$
$\frac{r_{1}}{r_{2}} = \frac{M_{S O_{2}}}{M}$
Thus, $\frac{18015}{120 / 20} = \sqrt{\frac{64}{M}}$
$2 = \sqrt{\frac{64}{M}}$
So, $M = 16$

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