Question

180mL of a hydrocarbon diffuses through a porous membrane in 15 minutes while 120mL of $S{{O}_{2}}$ under identical conditions diffuses in 20 minutes. What is the molecular mass of hydrocarbon?

Answer 1

Hint: Try to recall graham's law of diffusion. We need to apply graham's law in order to find out the molecular mass of the hydrocarbon.
Formula: $\dfrac{{{r}_{1}}}{{{r}_{2}}}=\dfrac{\sqrt{{{M}_{2}}}}{\sqrt{{{M}_{1}}}}$
Where,
${{r}_{1}}$ is the rate of diffusion of first gas,
${{r}_{2}}$ is the rate of diffusion of second gas,
${{M}_{1}}$ is the molar mass of first gas,
${{M}_{2}}$ is the molar mass of second gas.

Complete step by step answer:
We will try to understand Graham's law of diffusion and then apply the formula to find the answer.
-Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of its molecular weight.

$r\propto \dfrac{1}{M}$
-The complete explanation of the law was given by the kinetic theory of gases a few years later. Graham's law is used in isolating isotopes of an element as they have different rates of diffusion.
-Rate of diffusion (r) =$\dfrac{volume\ diffused}{time\ taken}$

We will now calculate rates of diffusion for the two gases in consideration.
${{r}_{1}}=\dfrac{180}{15}=12$ mL s-1
${{r}_{2}}=\dfrac{120}{20}=6$ mL s-1
Molecular mass of $S{{O}_{2}}({{M}_{2}})=64g.mo{{l}^{-1}}$
Substituting the values in the formula for graham's law of diffusion:
$\dfrac{12}{6}=\dfrac{\sqrt{64}}{\sqrt{{{M}_{1}}}}$
$2=\dfrac{8}{\sqrt{{{M}_{1}}}}$
${{M}_{1}}=16g$
Therefore, the molecular mass of the hydrocarbon is 16g.

Note: The formula mentioned above is used when the diffusion of gases happens at the same conditions like temperature, pressure etc. Remember that the rate of diffusion is inversely proportional to square root of molecular mass to avoid errors.