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(B) According to Graham's Law of diffusion,$\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$.Given,$\frac{r_{H_2}}{r_{C_n H_{2n-2}}} = 3\sqrt{3} = \sqrt{27}$

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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,$\frac{r_{H_2}}{r_{C_n H_{2n-2}}} = 3\sqrt{3} = \sqrt{27}$.Therefore,$\sqrt{\frac{M_{C_n H_{2n-2}}}{M_{H_2}}} = \sqrt{27}$.Since $M_{H_2} = 2 \ g/mol$,we have $\frac{M_{C_n H_{2n-2}}}{2} = 27$.$M_{C_n H_{2n-2}} = 54 \ g/mol$.The molar mass of $C_n H_{2n-2}$ is $12n + (2n - 2) = 14n - 2$.Equating the two,$14n - 2 = 54$,which gives $14n = 56$,so $n = 4$.

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

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