Question

A mixture of ${{C}_{2}}{{H}_{4}}$ and $C{{O}_{2}}$ effuses four times slowly as ${{H}_{2}}$ under identical conditions of temperature and pressure. The molar ratio of ${{C}_{2}}{{H}_{4}}$ to $C{{O}_{2}}$ in the mixture is
(A) 1:1
(B) 2:1
(C) 3:1
(D) 3:2

Answer 1

Hint: The answer to this question involves the concept of Graham’s law of diffusion/effusion which is inversely proportional to square root of molecular mass of gas and given as $r\propto \dfrac{1}{\sqrt{M}}$ and then calculating molecular weight of the mixture.

Complete step by step solution:
From the classes of physical chemistry, we have come across the concepts of diffusion of gases because they possess kinetic energy. Here, effusion refers to the movement of gas particles within a small hole.
- Therefore, on the basis of this concept rate of effusion was derived by Graham which is called Graham's law of effusion.
- According to this law, the rate of effusion of gas is inversely proportional to the molecular mass of gas which is given by the relation,
$r\propto \dfrac{1}{\sqrt{M}}$
-Now, effusion rate of hydrogen and rate of effusion of mixture can be denotes as ${{r}_{{{H}_{2}}}}$and ${{r}_{mix}}$ respectively.
According to the data given we have,
${{r}_{{{H}_{2}}}}=4\times {{r}_{mix}}$
Thus, the ratio of effusion rates of hydrogen gas and the mixture can be given as,
$\dfrac{{{r}_{{{H}_{2}}}}}{{{r}_{mix}}}=\sqrt{\dfrac{{{M}_{mix}}}{{{M}_{{{H}_{2}}}}}}$
By substituting values,
$4=\sqrt{\dfrac{{{M}_{mix}}}{2}}$ $\Rightarrow {{M}_{mix}}=32$
Now, ${{M}_{mix}}=\dfrac{{{M}_{1}}{{n}_{1}}+{{M}_{2}}{{n}_{2}}}{{{n}_{1}}+{{n}_{2}}}$
where, ${{M}_{1}}$= molecular weight of ${{C}_{2}}{{H}_{4}}$, ${{n}_{1}}$= number of moles of ${{C}_{2}}{{H}_{4}}$
${{M}_{2}}$= molecular weight of $C{{O}_{2}}$ and ${{n}_{2}}$= number of moles of $C{{O}_{2}}$
Therefore, by substituting the values,
$32=\dfrac{(28\times {{n}_{{{C}_{2}}{{H}_{4}}}})+(44\times {{n}_{C{{O}_{2}}}})}{{{n}_{{{C}_{2}}{{H}_{4}}}}+{{n}_{C{{O}_{2}}}}}$
$\Rightarrow 32{{n}_{{{C}_{2}}{{H}_{4}}}}+32{{n}_{C{{O}_{2}}}}=28{{n}_{{{C}_{2}}{{H}_{4}}}}+44{{n}_{C{{O}_{2}}}}$
Thus, we get $\dfrac{{{n}_{{{C}_{2}}{{H}_{4}}}}}{{{n}_{C{{O}_{2}}}}}=\dfrac{12}{4}=\dfrac{3}{1}$

Therefore, the ratio is 3:1 and the correct option is (C).

Note: When the questions are on the basis of effusion and molar ratio then Graham’s law is applicable and this point is to be noted and not to be confused with Fick’s law of diffusion which states that rate of diffusion is directly proportional to surface area and concentration and inversely proportional to thickness of membrane.