A hydrocarbon diffuses 3sqrt{3} times slower | vedclass

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(C) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $\frac{r_1}{r

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$C_4H_6$

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(C) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$.Given that the hydrocarbon $(r_h)$ diffuses $3\sqrt{3}$ times slower than hydrogen $(r_{H_2})$,we have $\frac{r_{H_2}}{r_h} = 3\sqrt{3}$.Substituting the molar masses ($M_{H_2} = 2 \ g/mol$ and $M_h$ for the hydrocarbon): $\frac{r_{H_2}}{r_h} = \sqrt{\frac{M_h}{M_{H_2}}} = 3\sqrt{3}$.Squaring both sides: $\frac{M_h}{2} = (3\sqrt{3})^2 = 9 	imes 3 = 27$.Therefore,$M_h = 27 	imes 2 = 54 \ g/mol$.Checking the molar masses of the options:$A: C_2H_4 = 2(12) + 4(1) = 28 \ g/mol$$B: C_3H_6 = 3(12) + 6(1) = 42 \ g/mol$$C: C_4H_6 = 4(12) + 6(1) = 48 + 6 = 54 \ g/mol$$D: C_6H_6 = 6(12) + 6(1) = 78 \ g/mol$.The hydrocarbon with molar mass $54 \ g/mol$ is $C_4H_6$.

$A$ hydrocarbon diffuses $3\sqrt{3}$ times slower than hydrogen gas. Identify the hydrocarbon.

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