Question

For \[50\]ml of hydrogen diffuses through a small hole in \[20\] minutes. The time taken for \[40\] ml of oxygen to diffuse under the same conditions in minutes is:
A.$12\min $
B.$64\min $
C.$8\min $
D.$32\min $

Answer 1

Hint: We know that the movement of particles which is gases or liquid from their higher concentration to lower concentration with any external disturbance as they internally collide with each other and eventually spread throughout the empty space. The formula to determine ratio of both rate we are going to use is \[{{\left( \dfrac{{{r}_{{{H}_{2}}}}}{{{r}_{{{O}_{2}}}}} \right)}^{2}}=\dfrac{{{M}_{{{O}_{2}}}}}{{{M}_{{{H}_{2}}}}}\]

Complete answer:
As we know, the process of diffusion is defined as the migration of some particles or anything from a high concentration area to a low concentration area. The gradient difference in the concentration contributes to diffusion. A gradient is defined as the variation present in the magnitude of a variable.
Diffusion is of two types. The simple diffusion in which the substance moves through the semipermeable membrane or without the help from transport protein. The other is facilitated diffusion which is a passive diffusion in which the molecules move from the high concentration to low concentration through the cell membrane. The factors affecting diffusion are temperature, particle size and area of interaction. In the given question we have asked to determine the time taken for diffusion as we already know that the rate of hydrogen gas is given by $Rate({{H}_{2}})=\dfrac{50}{20}=2.5$ similarly the rate of oxygen gas can be written as $Rate({{O}_{2}})=\dfrac{40}{t}.$
Now, we have to ratio both the rate;
\[{{\left( \dfrac{{{r}_{{{H}_{2}}}}}{{{r}_{{{O}_{2}}}}} \right)}^{2}}=\dfrac{{{M}_{{{O}_{2}}}}}{{{M}_{{{H}_{2}}}}}\] or \[\dfrac{{{r}_{{{H}_{2}}}}}{{{r}_{{{O}_{2}}}}}=\sqrt{\dfrac{{{M}_{{{O}_{2}}}}}{{{M}_{{{H}_{2}}}}}}\] here we know that molecular weight of oxygen is $16$ and of ${{O}_{2}}=2\times 16=32$ similarly molecular weight of hydrogen is one and ${{H}_{2}}=2\times 1=2.$
On substituting the values in the above equation, we get;
\[\dfrac{2.5}{\left( \dfrac{40}{t} \right)}=\sqrt{\dfrac{32}{2}}\] which implies \[\Rightarrow 2.5\times \dfrac{t}{40}=\sqrt{16}\]
On further solving we get;
\[\Rightarrow t=\dfrac{4\times 40}{2.5}=64\min .\]
Therefore, the correct answer is option B.

Note:
Remember that the distance between the particles is more in gases than liquids which results in fast diffusion in gases than liquids. So, the kinetic energy is more in gases particles so the diffusion in gases is quicker than in liquid.