Question

A teacher enters a classroom from the front door while a student from the back door. There are 13 equidistant rows of benches in the classroom. The teacher releases ${N_2}O$, the laughing gas, from the first bench while the student releases the weeping gas ($ {C_6}{H_{11}}OBr $) from the last bench. At which row will the students start laughing and weeping simultaneously?
A. 7
B. 10
C. 9
D. 8

Answer 1

Hint: Both laughing gas ($ {N_2}O $) and weeping gas ($ {C_6}{H_{11}}OBr$) when released in the classroom, will spread in the classroom by the process of diffusion. Diffusion of gases refers to the movement of the gaseous particles from the area of their high concentration to the area of their low concentration. Gaseous particles undergo diffusion due to their kinetic energy.

Formula used: According to Graham’s law of effusion,
$ dfrac{{rat{e_A}}}{{rat{e_B}}} = \sqrt {\dfrac{{{M_B}}}{{{M_A}}}} $
Here, $ rat{e_A}$ is the rate of diffusion of gas A and $ rat{e_B}$ is the rate of diffusion of gas B.
The $ {M_A}$ represents the molar mass of gas A and $ {M_B}$ is the molar mass of gas B.

Complete step by step answer:
Here, two gases were released in the classroom from two different sides of the classroom. There are 13 rows of benches situated at equal distance from each other in the classroom. Now, we have to find out at which row the laughing gas and weeping gas will reach at the same time.
For that we need to use Graham’s law. It states that, the rate of diffusion of a gas is inversely proportional to that of the molar mass of that gas. Mathematically, it can be expressed as:
$ rate \propto \sqrt {\dfrac{1}{M}}$
Let us assume, $ {R_1}$ is the row from the front and ${R_2}$ is the row from the last of the classroom from which students will start laughing and weeping simultaneously. The molar mass of the laughing gas released from front is ${M_1}$. The molar mass of the weeping gas released from the last bench is ${M_2}$.
Based on this information, on applying Graham’s law we get,
$\dfrac{{{R_1}}}{{{R_2}}} = \sqrt {\frac{{{M_2}}}{{{M_1}}}} $
Using this formula, we can do further calculations.
Step 1: Now, we know the chemical formula of both laughing gas and weeping gas, so we can find out values of both ${M_1}$ and ${M_2}$.
$Molar mass of compound = Molecular weight of compound$
The molar mass of laughing gas $({N_2}O)$ is:
$Molar\;mass\;of\;{N_2}O = 2 \times atomic\;weight\;of\;nitrogen + 1 \times atomic\;weight\;of\;oxygen $
$Molar\; Mass\; Of\;{N_2}O = 2 \times 14 + 1 \times 16 = 28 + 16 $
$\therefore Molar\;mass\;of\;{N_2}O = 44$
Similarly, the molar mass of weeping gas $({C_6}{H_{11}}OBr) $ is:
$Molar\;mass\;of\;{C_6}{H_{11}}OBr = 6 \times 12 + 11 \times 1 + 1 \times 16 + 1 \times 80$
$\therefore Molar\;mass\;of\;{C_6}{H_{11}}OBr = 179$
The molar mass of laughing gas is 44 and that of weeping gas is 179.
Step 2: Let us put the calculated values of molar masses of both laughing and weeping gas in the formula obtained from Graham’s law.
$\dfrac{{{R_1}}}{{{R_2}}} = \sqrt {\dfrac{{179}}{{44}}} $
$\therefore \dfrac{{{R_1}}}{{{R_2}}} = 2.017$
$\therefore \dfrac{{{R_1}}}{{{R_2}}} = 2$
For simplification, we will round up the figure 2.017 as 2.
Step 3: We have to find out the number of the row at which the student will start laughing and weeping simultaneously.
Let us assume the number of rows from the front at which student will start laughing and weeping simultaneously as x.
Therefore, ${R_1} = x$
We know that, $ {R_1} + {R_2} = 13 $
On putting x as the value of $ {R_1}$ in the above equation, we get,
$ {R_2} = 13 - x $
Now, we have got the values for $ {R_1}$ and $ {R_2}$, substitute these values in the given formula,
$ \dfrac{{{R_1}}}{{{R_2}}} = 2 $
$ \therefore \dfrac{x}{{(13 - x)}} = 2 $
$ \therefore x = 2 \times (13 - x) $
$ \therefore x = 26 - 2x $
$ \therefore 3x = 26 $
$ \therefore x = \dfrac{{26}}{3} = 8.67 \approx 9 $
The student will start laughing and weeping simultaneously at 9 rows.

So, the correct answer is Option B.

Note: The Graham’s law of diffusion helps in determining the molecular weight of an unknown gas from its rate of diffusion. It is also useful when the gases with different densities need to be separated. Isotopes of some elements can be separated using Graham’s law of diffusion.
Students should not get confused between the term ‘effusion’ and ‘diffusion’. The term effusion refers to the movement of gas molecules through a microscopic opening. The collision between effusing gas particles is negligible. Graham’s law is also applicable to the effusion of gases.
$ \dfrac{{rat{e_A}}}{{rat{e_B}}} = \sqrt {\dfrac{{{M_B}}}{{{M_A}}}} $
Here, $ rat{e_A}$ is the rate of effusion of gas A and $ rat{e_B}$ is the rate of effusion of gas B.
The $ {M_A}$ represents the molar mass of gas A and $ {M_B}$ is the molar mass of gas B.