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180ml of a hydrocarbon diffuses through a porous membrane in 15 minutes while 120ml of SO_{2 }under similar conditions diffuses in 20 minutes. What is the molecular mass of hydrocarbon? and how

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

$r=\frac{Volumeeffused}{time}=\frac{V}{t}$

So,

$\frac{{r}_{1}}{{r}_{2}}=\sqrt{\frac{{M}_{2}}{{M}_{1}}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{V}_{1}/{t}_{1}}{{V}_{2}/{t}_{2}}=\sqrt{\frac{{M}_{2}}{{M}_{1}}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{180/15}{120/20}=\sqrt{\frac{64}{{M}_{1}}}(MolecularmassofS{O}_{2}=64g/mol)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{180\times 20}{120\times 15}=\frac{8}{\sqrt{{M}_{1}}}\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{{M}_{1}}=\frac{8\times 120\times 15}{180\times 20}\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{{M}_{1}}=4\phantom{\rule{0ex}{0ex}}Onsquaringbothsides\phantom{\rule{0ex}{0ex}}\Rightarrow {M}_{1}=16$

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

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