At identical temperature and pressure,the rat | vedclass

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(B) According to Graham's Law of Diffusion,$\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$.Given $\frac{r_{H_2}}{r_{hydrocarbon}} = 3\sqrt{3}$,where $M_{H_

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$4$

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(B) According to Graham's Law of Diffusion,$\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$.Given $\frac{r_{H_2}}{r_{hydrocarbon}} = 3\sqrt{3}$,where $M_{H_2} = 2 \ g/mol$.So,$3\sqrt{3} = \sqrt{\frac{M_{hydrocarbon}}{2}}$.Squaring both sides: $(3\sqrt{3})^2 = \frac{M_{hydrocarbon}}{2} \Rightarrow 27 = \frac{M_{hydrocarbon}}{2}$.$M_{hydrocarbon} = 54 \ g/mol$.The molecular formula is $C_nH_{2n-2}$,so $12n + (2n-2) = 54$.$14n - 2 = 54$ $\Rightarrow 14n = 56$ $\Rightarrow n = 4$.

At identical temperature and pressure,the rate of diffusion of hydrogen gas is $3\sqrt{3}$ times that of a gaseous hydrocarbon having molecular formula $C_nH_{2n-2}$. What is the value of $n$?

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