Question

A gaseous mixture comprising of equal moles of $ {{H}_{2}} $ , $ {{O}_{2}} $ and $ M\text{ (mass = 128)} $ was subjected to series of effusion steps. What will be the number of effusion steps required so as to change the composition to: one in which will be $ \text{lightest : heaviest}\, $ is $ 4096:1 $ . What will be the composition of the mixture (w.r.t all the gases)?
A. $ 4,4096:16:1 $
B. $ 6,4096\times 64:64:1 $
C. $ 4,2048:8:1 $
D. $ 5,4096:16:1 $

Answer 1

Hint: Graham’s law of diffusion: It states that the rate of effusion or diffusion of a gas is inversely proportional to the square root of the molecular mass of the gas. In simple words, it explains that the heavier the mass of a gas, the slower it will displace from one point to another.

Complete answer:
According to the question, the given data is as follows:
 $ \dfrac{\text{lightest}}{\text{heaviest}}=\dfrac{4096}{1}\,\,-(i) $
Now, it is given that a gaseous mixture of $ {{H}_{2}} $ , $ {{O}_{2}} $ and $ M $ comprises an equal number of moles in the mixture. So, the molecular mass of each gas is as follows:
 $ {{H}_{2}}=2\,gmo{{l}^{-1}} $
 $ {{O}_{2}}=32\text{ }gmo{{l}^{-1}} $
 $ M=128\text{ }gmo{{l}^{-1}} $
Therefore, according to graham’s law the ratio for rate of effusion can be written as follows:
 $ {{r}_{{{H}_{2}}}}:{{r}_{{{O}_{2}}}}:{{r}_{M}}=\dfrac{1}{\sqrt{{{M}_{{{H}_{2}}}}}}:\dfrac{1}{\sqrt{{{M}_{{{O}_{2}}}}}}:\dfrac{1}{\sqrt{{{M}_{M}}}} $
 $ \Rightarrow {{r}_{{{H}_{2}}}}:{{r}_{{{O}_{2}}}}:{{r}_{M}}=\dfrac{1}{\sqrt{2}}:\dfrac{1}{\sqrt{32}}:\dfrac{1}{\sqrt{128}} $
Because, $ {{H}_{2}} $ is the lightest gas and $ M $ is the heaviest gas, so the expression for the rate of effusion can be written as follows:
 $ \dfrac{{{r}_{{{H}_{2}}}}}{{{r}_{M}}}=\sqrt{\dfrac{128}{2}} $
But this is the expression for the rate of diffusion at the first step. For N steps of diffusion, the expression will be as follows:
 $ \dfrac{{{r}_{{{H}_{2}}}}}{{{r}_{M}}}={{\left( \sqrt{\dfrac{128}{2}} \right)}^{N}} $
Substituting value from equation (i)-
 $ {{\left( \sqrt{\dfrac{128}{2}} \right)}^{N}}=\dfrac{4096}{1} $
 $ \Rightarrow {{2}^{3N}}={{2}^{12}} $
As the base is same, so we can compare the powers, therefore, the value of N is as follows:
 $ 3N=12 $
 $ \Rightarrow N=4 $
Hence, the number of effusion steps required to change the composition of mixture $ =4 $ .
Thus, the composition of the mixture after four steps of effusion will be as follows:
 $ {{r}_{{{H}_{2}}}}:{{r}_{{{O}_{2}}}}:{{r}_{M}}={{\left( \dfrac{1}{\sqrt{2}} \right)}^{4}}:{{\left( \dfrac{1}{\sqrt{32}} \right)}^{4}}:{{\left( \dfrac{1}{\sqrt{128}} \right)}^{4}} $
 $ \Rightarrow {{r}_{{{H}_{2}}}}:{{r}_{{{O}_{2}}}}:{{r}_{M}}=\dfrac{1}{4}:\dfrac{1}{1024}:\dfrac{1}{16384} $
On simplifying ratios:
 $ \Rightarrow {{r}_{{{H}_{2}}}}:{{r}_{{{O}_{2}}}}:{{r}_{M}}=4096:16:1 $
Hence, option (A) i.e., $ 4,4096:16:1 $ is the correct answer.

Note:
Do not get confused with the terms diffusion and effusion. Diffusion occurs when gases mixed through random motion which results in collision of gaseous molecules whereas effusion is the tendency of a gas to escape from a small (pinhole) opening. Graham’s law is applicable for both diffusion as well as effusion.