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

The correct option is A $1$

We will use the Graham’s Law of diffusion to solve this question which is given by:

$\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}$ where $M_1$ and $M_2$ are molar masses of the given gas.

Given, rate of diffusion of $H_2$ is $3\sqrt{3}$ times the rate of diffusion of $C_n{H}_{2n-2}$ .

Molar mass of $H_2$ is $2$ .

Substituting all the values in formula

$\frac{r_{H_2}}{r_{C_n{H}_{2n-2}}}=\sqrt{\frac{M_{{}_{C_n{H}_{2n-2}}}}{M_{H_2}}}$

Let, rate of diffusion of $H_2$ be $3\sqrt{3}x$ and rate of diffusion of $C_n{H}_{2n-2}$ be $x.$

$\Rightarrow \frac{3\sqrt{3}x}{x}=\sqrt{\frac{M_{{}_{C_n{H}_{2n-2}}}}{M_{H_2}}}$

Taking square on both side

$\Rightarrow 27=\frac{M_{{}_{C_n{H}_{2n-2}}}}{M_{H_2}}$

Molar mass of $H_2$ is $2$ .

$\Rightarrow 27=\frac{M_{{}_{C_n{H}_{2n-2}}}}{2}$

$\Rightarrow 54=M_{{}_{C_n{H}_{2n-2}}}$

Now, the compound which has molar mass of $54$ is $C_4{H}_6$.

Threrefore the value of $n=4$.

Option B is correct.