Question

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 $'n'$ ?

• A. 1
• B. 4
• C. 3
• D. 8

Answer 1

Answer: (B)

4

Answer 2

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