Question

At the identical temperature and pressure, the rate of diffusion of hydrogen gas is $\sqrt[3]{33}$ times that of a hydrocarbon having molecular formula ${C_{n}H}_{2n-2}$. What is the volume of $n$?

• A. 1
• B. 4
• C. 3
• D. 8

Answer 1

The correct option is B 4

Given, rate of diffusion of hydrogen $=r_{H_2}$

rate of diffusion of hydrocarbon $=r_{HC}$

Also $r_{H_2}$$=3\sqrt{3}{r}_{HC}⟶\left(1\right)$

Also P and T is constant.

We know by graham's law of diffusion

$r\propto \frac{1}{\sqrt{M}}$ M is molar mass

$\frac{r_{H_2}}{r_{HC}}=\sqrt{\frac{M_{HC}}{M_{H_2}}}$

$\frac{r_{H_2}}{r_{HC}}=\sqrt{\frac{M_{HC}}{2}}$

From 1st equation,

$3\sqrt{3}=\sqrt{\frac{M_{HC}}{2}}$ [squaring on both sides]

$27=M_{HC}/2$

$M_{HC}=54$

Molar mass of hydrocarbon $C_n{H}_{2n-2}=54g/mole$

We know

$1C=12g/mole,1H=1g/mole$

$12\times n+1\times (2n-2)=54$

$n=4$