Question

Sulfur dioxide and oxygen were allowed to diffuse through a porous partition. $20\text{ }d{{m}^{3}}$ of $S{{O}_{2}}$ diffuse through the porous partition in 60 sec. The volume of ${{O}_{2}}$ in $d{{m}^{3}}$ which diffuses under the similar condition in 30sec will be: (Atomic mass of sulfur 32 u)
(A) 7.09
(B) 14.1
(C) 10.0
(D) 28.2

Answer 1

Hint: The rate of diffusion is directly proportional to the ratio of volume to the time taken and the rate of the diffusion is inversely proportional to the square root of the atomic mass of the element. Combine these two we can solve the volume of oxygen.

Complete step by step solution:
The rate of the diffusion or effusion is directly proportional to the ratio of the volume to the time taken. This is written as:
$r\propto \dfrac{V}{t}$
 So, the rate of diffusion of sulfur dioxide is directly proportional to the ratio of the volume of sulfur dioxide to the time taken of sulfur dioxide.
${{r}_{S{{O}_{2}}}}\propto \dfrac{{{V}_{S{{O}_{2}}}}}{{{t}_{S{{O}_{2}}}}}$
And the rate of diffusion of oxygen is directly proportional to the ratio of the volume of oxygen to the time taken of oxygen.
${{r}_{{{O}_{2}}}}\propto \dfrac{{{V}_{{{O}_{2}}}}}{{{t}_{{{O}_{2}}}}}$
 Rate of the diffusion is inversely proportional to the square root of the atomic mass of the element.
$r\propto \dfrac{1}{\sqrt{M}}$
 Rate of the diffusion of sulfur dioxide is inversely proportional to the square root of the atomic mass of the sulfur dioxide.
${{r}_{S{{O}_{2}}}}\propto \dfrac{1}{\sqrt{{{M}_{S{{O}_{2}}}}}}$
 Rate of the diffusion of oxygen is inversely proportional to the square root of the atomic mass of the oxygen.
${{r}_{{{O}_{2}}}}\propto \dfrac{1}{\sqrt{{{M}_{{{O}_{2}}}}}}$
So combining the above relation can be used to derive the volume of the oxygen.
$\dfrac{{{r}_{{{O}_{2}}}}}{{{r}_{S{{O}_{2}}}}}=\dfrac{\dfrac{{{V}_{{{O}_{2}}}}}{{{t}_{{{O}_{2}}}}}}{\dfrac{{{V}_{S{{O}_{2}}}}}{{{t}_{S{{O}_{2}}}}}}$
$\dfrac{{{r}_{{{O}_{2}}}}}{{{r}_{S{{O}_{2}}}}}=\dfrac{{{V}_{{{O}_{2}}}}}{{{V}_{S{{O}_{2}}}}} \times \text{ }\dfrac{{{t}_{S{{O}_{2}}}}}{{{t}_{{{O}_{2}}}}}$
$r\propto \dfrac{\sqrt{{{M}_{S{{O}_{2}}}}}}{\sqrt{{{M}_{{{O}_{2}}}}}}$
Equating both, we get
$\dfrac{{{V}_{{{O}_{2}}}}}{{{V}_{S{{O}_{2}}}}} \times \text{ }\dfrac{{{t}_{S{{O}_{2}}}}}{{{t}_{{{O}_{2}}}}}=\dfrac{\sqrt{{{M}_{S{{O}_{2}}}}}}{\sqrt{{{M}_{{{O}_{2}}}}}}$
${{V}_{{{O}_{2}}}}=\dfrac{\sqrt{{{M}_{S{{O}_{2}}}}}}{\sqrt{{{M}_{{{O}_{2}}}}}} \times \text{ }\dfrac{{{t}_{{{O}_{2}}}}}{{{t}_{S{{O}_{2}}}}} \times \text{ }{{V}_{S{{O}_{2}}}}$
So, the values from the question can be applied on this equation, we get
${{V}_{{{O}_{2}}}}=\dfrac{\sqrt{32+32}}{\sqrt{32}} \times \text{ }\dfrac{30}{60} \times \text{ 20}$
${{V}_{{{O}_{2}}}}=10\sqrt{2}=14.1\text{ }d{{m}^{3}}$

Therefore, the correct answer is an option (B)- 14.1.

Note: The rate of diffusion or effusion is also inversely proportional to the densities of the element and this helps to separate the gases having different densities. These formulas can also be used to separate isotopes.