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(B) According to Graham's Law of diffusion,$\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$.Given $\frac{r_{H_2}}{r_{gas}} = 3\sqrt{3}$,where $M_{H_2} = 2 \

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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 $\frac{r_{H_2}}{r_{gas}} = 3\sqrt{3}$,where $M_{H_2} = 2 \ g/mol$.So,$3\sqrt{3} = \sqrt{\frac{M_{gas}}{2}}$.Squaring both sides,$(3\sqrt{3})^2 = \frac{M_{gas}}{2} \implies 9 	imes 3 = \frac{M_{gas}}{2} \implies 27 = \frac{M_{gas}}{2}$.Thus,$M_{gas} = 54 \ g/mol$.The molecular formula is $C_n H_{2n-2}$,so $12n + (2n - 2) = 54$.$14n - 2 = 54 \implies 14n = 56 \implies n = 4$.

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 $C_n H_{2n-2}$. What is the value of $n$?

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