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$$
\text { At identical temperature and pressure, the rate of diffusion of hydrogen gas is } 3 \sqrt{3} \text { times that }
$$
of a hydrocarbon having molecular formula $\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}-2}$. What is the value of ' $n$ '?
Options:
\text { At identical temperature and pressure, the rate of diffusion of hydrogen gas is } 3 \sqrt{3} \text { times that }
$$
of a hydrocarbon having molecular formula $\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}-2}$. What is the value of ' $n$ '?
Solution:
2175 Upvotes
Verified Answer
The correct answer is:
4
$$
\begin{aligned}
& \text { Hints : } \frac{r_{H_2}}{r_{C_n H_{2 n-2}}}=\sqrt{\frac{M_{\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}-2}}}{\mathbf{M}_{\mathrm{H}_2}}}=\sqrt{\frac{\mathbf{M}_{\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}-2}}}{2}} \\
& \because \sqrt{\frac{\mathrm{M}_{\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}-2}}}{2}}=3 \sqrt{3}=\sqrt{27} \\
& \Rightarrow \mathbf{M}_{\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}-2}}=27 \times 2=54
\end{aligned}
$$
Hence, $12 n+(2 n-2) \times 1=54 \Rightarrow 14 n=56 \Rightarrow n=4$ Thus Hydrocarbon is $\mathrm{C}_4 \mathrm{H}_6$
\begin{aligned}
& \text { Hints : } \frac{r_{H_2}}{r_{C_n H_{2 n-2}}}=\sqrt{\frac{M_{\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}-2}}}{\mathbf{M}_{\mathrm{H}_2}}}=\sqrt{\frac{\mathbf{M}_{\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}-2}}}{2}} \\
& \because \sqrt{\frac{\mathrm{M}_{\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}-2}}}{2}}=3 \sqrt{3}=\sqrt{27} \\
& \Rightarrow \mathbf{M}_{\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}-2}}=27 \times 2=54
\end{aligned}
$$
Hence, $12 n+(2 n-2) \times 1=54 \Rightarrow 14 n=56 \Rightarrow n=4$ Thus Hydrocarbon is $\mathrm{C}_4 \mathrm{H}_6$
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