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 $'n'$ ?
$\frac{r_{ H _{2}}}{r_{ C _{n} H _{2 n-2}}}=\sqrt{\frac{{M}_{ C _{n} H _{2 n-2}}}{M_{ H _{2}}}}$ $=\sqrt{\frac{M_{ C _{n} H _{2 n-2}}}{2}}$ $\because \sqrt{\frac{M_{ C _{n} H _{2 n-2}}}{2}}=3 \sqrt{3}=\sqrt{27}$ $\Rightarrow M_{C_{n} H_{2 n-2}}=27 \times 2=54$ Hence, $12 n+(2 n-2) \times 1=54$ $\Rightarrow 14 n=56 $ $\Rightarrow n=4$ Thus, hydrocarbon is $C _{4} H _{6}$.