Question

$N{H_3}$ and $HCl$ gas are introduced simultaneously from the two ends of a long tube. A white ring of $N{H_4}Cl$ appears first:
A.nearer to the $HCl$ end
B.at the centre of the tube
C.throughout the tube
D.nearer to the $N{H_3}$ end

Answer 1

First we must know that the rate of diffusion is inversely proportional to the molecular mass of the gas. Because, two gas $N{H_3}$ and $HCl$are diffusing with each other and forming $N{H_4}Cl$ white ring.

Complete step by step answer:
First we will discuss what all are the possibilities in this question.
Let’s assume we have two gasses A and B which react to give a ring (just like the question) now in case 1, if the gas A diffuses faster than gas B then the ring will form near to B. In case 2, if the rate of diffusion is same for both A and B then the ring will be formed in the centre. And In case 3, if the gas B diffuses faster than A then the ring will be formed near A.
Now, the diffusibility of a gas is inversely proportional to the molecular mass of the gas.
In case of$HCl$,
The molecular mass of the gas will be 1 (atomic mass of Hydrogen) + 35.5 (atomic mass of chlorine) = $36.5{\text{ }}g/mol$(molecular mass of $HCl$).
Now in case of NH3,
The molecular mass will be 14 (atomic mass of Nitrogen) + 3 X 1 (atomic mass of hydrogen is 1 and there are 3 hydrogen’s) = $17{\text{ }}g/mol$(molecular mass of$N{H_3}$).
Since the molecular mass of $HCl$ is more than $N{H_3}$, this means that the diffusibility of $N{H_3}$ is greater than$HCl$. Hence, the White ring will appear nearer to the $HCl$.
So, the answer to this question A i.e nearer to the $HCl$ end.

Note:
We must know that the concept of diffusibility helps us determine how fast a gas reacts with other gas as well as tell us about the speed at which it moves from one place to another. A famous example for the same is diffusion of perfume, when one spray a perfume in one corner the odour of perfume reaches the other end of the room in seconds. This can be explained using the concept of diffusibility.