360 cm^3 of a hydrocarbon diffuses in 30 min | vedclass

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(A) According to Graham's Law of diffusion: $\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$Here, $r_1 = \frac{360}{30} = 12 \ cm^3/min$ (for hydrocarbon) a

Vedclass · Accepted Answer

$CH_4$

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(A) According to Graham's Law of diffusion: $\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$Here, $r_1 = \frac{360}{30} = 12 \ cm^3/min$ (for hydrocarbon) and $r_2 = \frac{360}{60} = 6 \ cm^3/min$ (for $SO_2$).$M_2$ (molecular weight of $SO_2$) = $64 \ g/mol$.Substituting the values: $\frac{12}{6} = \sqrt{\frac{64}{M_1}}$$2 = \sqrt{\frac{64}{M_1}}$Squaring both sides: $4 = \frac{64}{M_1}$$M_1 = \frac{64}{4} = 16 \ g/mol$.The hydrocarbon with a molecular weight of $16 \ g/mol$ is $CH_4$.

$360 \ cm^3$ of a hydrocarbon diffuses in $30$ minutes, while under the same conditions $360 \ cm^3$ of $SO_2$ gas diffuses in one hour. The molecular formula of the hydrocarbon is

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