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(B) According to Graham's Law of diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $r \propto \f

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$2$

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(B) According to Graham's Law of diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $r \propto \frac{1}{\sqrt{M}}$.Therefore,$\frac{r_X}{r_{	ext{gas}}} = \sqrt{\frac{M_{	ext{gas}}}{M_X}}$.Given that $r_X = 3 \sqrt{3} 	imes r_{	ext{gas}}$,we have $\frac{r_X}{r_{	ext{gas}}} = 3 \sqrt{3}$.Substituting the values: $3 \sqrt{3} = \sqrt{\frac{54}{M_X}}$.Squaring both sides: $(3 \sqrt{3})^2 = \frac{54}{M_X}$.$27 = \frac{54}{M_X}$.$M_X = \frac{54}{27} = 2 \ g \ mol^{-1}$.

At the same temperature and pressure,the rate of diffusion of gas $X$ is $3 \sqrt{3}$ times that of a gaseous hydrocarbon of molar mass $54 \ g \ mol^{-1}$. The molar mass of $X$ in $g \ mol^{-1}$ is

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