Question

A hydrocarbon gas diffuses through a porous plug diffuses at the $\dfrac{1}{6}th$ rate of diffusion of ${{\text{H}}_{\text{2}}}$ gas under similar conditions of P and T. The gas is:
A. ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}$
B. ${{\text{C}}_{\text{5}}}{{\text{H}}_{{\text{12}}}}$
C. ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}$
D. ${{\text{C}}_{\text{4}}}{\text{ }}{{\text{H}}_{{\text{10}}}}$

Answer 1

From the Graham’s law of diffusion, determine the molecular weight of hydrocarbon gas. Compare this molecular weight with the molecular weights of the compounds given in the options.

Complete answer:
According to the Graham's law of diffusion
$\dfrac{{{{\text{r}}_{\text{g}}}}}{{{{\text{r}}_{\text{H}}}}}{\text{ = }}\sqrt {\dfrac{{{{\text{M}}_{\text{H}}}}}{{{{\text{M}}_{\text{g}}}}}}$
Here, ${{\text{r}}_{\text{g}}}$ and ${{\text{r}}_{\text{H}}}$ are the rates of diffusion of the hydrocarbon gas and hydrogen gas respectively. Also ${{\text{M}}_{\text{g}}}$ and ${{\text{M}}_{\text{H}}}$ are the rates of diffusion of the hydrocarbon gas and hydrogen gas respectively.
Substitute values in the above expression
$\Rightarrow \dfrac{{\text{1}}}{{\text{6}}}{\text{ = }}\sqrt {\dfrac{{\text{2}}}{{{{\text{M}}_{\text{g}}}}}}$
Take square on both sides of the equation
$\Rightarrow \dfrac{{\text{1}}}{{{\text{36}}}} = \dfrac{{\text{2}}}{{{{\text{M}}_{\text{g}}}}}$
Rearrange the above equation and calculate the molecular weight of hydrocarbon gas
$\Rightarrow {{\text{M}}_{\text{g}}}{\text{ = 2}}\times \text{36}$
$\Rightarrow {{\text{M}}_{\text{g}}}{\text{ = 72}}$
The atomic masses (in grams per mole) of carbon and hydrogen are 12 and 1 respectively.
Calculate the molecular weights of the compounds given in the options
For ethane molecule, ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}} = 2\left( {12} \right) + 6\left( 1 \right) = 24 + 6 = 30$
For pentane molecule, ${{\text{C}}_{\text{5}}}{{\text{H}}_{{\text{12}}}} = 5\left( {12} \right) + 12\left( 1 \right) = 60 + 12 = 72$
For acetylene molecule, ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}} = 2\left( {12} \right) + 2\left( 1 \right) = 24 + 2 = 26$
For butane molecule, ${{\text{C}}_{\text{4}}}{\text{ }}{{\text{H}}_{{\text{10}}}} = 4\left( {12} \right) + 10\left( 1 \right) = 48 + 10 = 58$
The molecular weight of ${{\text{C}}_{\text{5}}}{{\text{H}}_{{\text{12}}}}$ matches with the value calculated from the Graham’s law of diffusion.

**Hence, the correct option is the option (B) ${{\text{C}}_{\text{5}}}{{\text{H}}_{{\text{12}}}}$

Note:**
According to Graham’s law of diffusion, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. The rate of diffusion is the ratio of the amount of gas diffused to the time needed for the diffusion.