Question

A specific volume of ${H_2}$​ requires $24\, s$ to diffuse out of a container. The time required by an equal volume of ${O_2}$​ to diffuse out under identical conditions, is:
A. $24\, s$
B. $96\, s$
C. $384\, s$
D. $192\,s$

Answer 1

The ratio of molar mass of oxygen to molar mass of the hydrogen is directly proportional to the ratio of square of the volume of the hydrogen to the square of the volume of oxygen. Also, the ratio of molar mass of oxygen to molar mass of the hydrogen is directly proportional to the ratio of square of the time taken by oxygen to diffuse out to the square of the time taken by hydrogen to diffuse out.

Complete step by step answer:
Given:Time required for ${H_2}$ to diffuse out of a container = ${T_H} = 24\,s$
Volume of oxygen =${V_O}$= Volume of Hydrogen =${V_H}$
Molecular mass of the ${H_2}$= Atomic mass of $H \times 2$
Atomic mass of $H = 1$
Hence, Molecular mass of the ${H_2}$ = $1 \times 2 = 2$
The molar mass of the oxygen is = Atomic mass of $O \times 2$
The molar mass of the oxygen is = $16 \times 2$
The molar mass of the oxygen is = $32$
We know that the ratio of molar mass of oxygen to molar mass of the hydrogen is directly proportional to the ratio of square of the volume of the hydrogen to the square of the volume of oxygen.
$\dfrac{{{M_{O}}}}{{{M_H}}}\propto \dfrac{{V_H^2}}{{V_O^2}}$……………………………………………………………………………………………………………………………………..………. (I)
Where,
Molar mass of oxygen=${M_O}$
Molar mass of Hydrogen =${M_H}$
Also, the ratio of molar mass of oxygen to molar mass of the hydrogen is directly proportional to the ratio of square of the time taken by oxygen to diffuse out to the square of the time taken by hydrogen to diffuse out.
$\dfrac{{{M_O}}}{{{M_H}}}\propto \dfrac{{T_O^2}}{{T_H^2}}$…………………………………………………………………………………………………………………………………………… (II)
Where Time required to diffuse oxygen out of the container = ${T_O}$
From equation (I) and (II),
$\Rightarrow \dfrac{{{M_O}}}{{{M_H}}} = \dfrac{{T_O^2}}{{T_H^2}} \times \dfrac{{V_H^2}}{{V_O^2}}$
Hence,
$\Rightarrow \sqrt {\dfrac{{{M_O}}}{{{M_H}}}} = \dfrac{{T_O^{}}}{{T_H^{}}} \times \dfrac{{V_H^{}}}{{V_O^{}}}$
$\Rightarrow \sqrt {16} = \dfrac{{T_O^{}}}{{24}}$
$\Rightarrow \sqrt {16} \times 24 = {T_O}$
$\Rightarrow {T_O} = 96\,s$

**Hence, option (B) is the correct answer.

Note: **
Molar mass of the gas depends on time required to diffuse out from the cylinder and volume of that cylinder. The above relations have been found by Thomas Graham in his famous Graham's Law of Diffusion to find the molecular mass of a different gas using it’s diffusion rate.