Question

180 mL of 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 the hydrocarbon?
(A) 8
(B) 16
(C) 24
(D) 32

Answer 1

Hint: This question is based on the concept of Graham’s law of diffusion, which states that the rate of diffusion of a gas is inversely proportional to the molecular weight of the gas. Diffusion is defined as the movement of a gas from a region of higher concentration to a region of lower concentration.

Complete step by step answer:
The mathematical representation of Graham’s law of diffusion is as follows:
\[Rate \propto \dfrac{1}{{\sqrt {Molecular\;Weight} }}\]
Or, \[Rate = k.\dfrac{1}{{\sqrt {Molecular\;Weight} }}\]
Where \[k\]is the proportionality constant.
Let the rate of diffusion of hydrocarbon be taken as \[{R_1}\] and that of \[S{O_2}\]being taken as \[{R_2}\].
Let the molecular weight of the unknown hydrocarbon be taken as \[x\].
According to Graham’s law of diffusion, the rate of diffusion of hydrocarbon will be:
\[{R_1} = \dfrac{k}{{\sqrt x }}\]
The molecular weight of \[S{O_2}\]can be calculated as follows: \[Atomic\;weight\;of\;S + 2 \times Atomic\;weight\;of\;O\]
\[ = 32\;g + 2 \times 16\;g\]
\[ = 32\;g + 32\;g\]
\[ = 64\;g\]
Therefore, the equation for the rate of diffusion of \[S{O_2}\]can be written as:
\[{R_2} = \dfrac{k}{{\sqrt {64\;g} }}\]
Dividing \[{R_1}\]by \[{R_2}\], we get;
\[\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{\dfrac{k}{{\sqrt x }}}}{{\dfrac{k}{{\sqrt {64\;g} }}}}\]
\[ \Rightarrow \dfrac{{{R_1}}}{{{R_2}}} = \dfrac{k}{{\sqrt x }} \times \dfrac{{\sqrt {64\;g} }}{k}\]
\[ \Rightarrow \dfrac{{{R_1}}}{{{R_2}}} = \sqrt {\dfrac{{64\;g}}{x}} {{ }} \to {{(i)}}\]
Now, the rate of diffusion of a gas is given by the following relation: \[Rate\;of\;diffusion = \dfrac{{Volume\;of\;gas\;diffused}}{{Time\;taken\;for\;diffusion}}\]
Given, The volume of hydrocarbon diffused \[ = 180\;ml\]
Time taken for diffusion \[ = 15\;minutes\]
The volume of $S{O_2}$ diffused \[ = 120\;ml\]
Time taken for diffusion \[ = 20\;minutes\]
Substituting the given values, we get the rate of diffusion of hydrocarbon, \[{R_1}\], as:
\[{R_1} = \dfrac{{180\;ml}}{{15\;mins.}}\]
\[ \Rightarrow {R_1} = 12\;ml/min\]
Similarly, for \[S{O_2}\], we get the rate of diffusion, \[{R_2}\], as follows:
\[{R_2} = \dfrac{{120\;ml}}{{20\;mins.}}\]
\[ \Rightarrow {R_2} = 6\;ml/min\]
Now, putting the values of \[{R_1}\]and \[{R_2}\]in equation \[(i)\], we get;
\[ \Rightarrow \dfrac{{12\;ml/min}}{{6\;ml/min}} = \sqrt {\dfrac{{64\;g}}{x}} \]
\[ \Rightarrow 2 = \sqrt {\dfrac{{64\;g}}{x}} \]
Squaring on both sides
\[ \Rightarrow {2^2} = \dfrac{{64\;g}}{x}\]
\[ \Rightarrow 4 = \dfrac{{64\;g}}{x}\]
On cross multiplication
\[ \Rightarrow 4x = 64\;g\]
\[ \Rightarrow x = \dfrac{{64\;g}}{4}\]
\[ \Rightarrow x = 16\;g\]
Therefore, the molecular weight of the given hydrocarbon is \[16g\].

So, the correct answer is Option B.

Note: The solution can be directly started with step \[\dfrac{{{R_1}}}{{{R_2}}} = \sqrt {\dfrac{{64\;g}}{x}} \] since we know that the rates of diffusion of two gases are inversely proportional to their molecular weights, according to Graham’s law of diffusion.
Graham’s law of diffusion is stated as:
\[Rate \propto \dfrac{1}{{\sqrt {Density} }}\]
Now, density is given as mass per unit volume which can be written as:
\[Density = \dfrac{{Mass}}{{Volume}}\]
If the temperature and pressure conditions are kept similar, then the number of moles of different gases will be the same for different volumes (Avogadro’s law). This tells us that the density of a gas is directly proportional to the molecular weight of a gas. Since density directly depends upon the molecular weight of a gas, Graham’s law of diffusion is written as:
\[Rate \propto \dfrac{1}{{\sqrt {Molecular\;Weight} }}\].