Question

An unknown gas ‘X’ has a rate of diffusion measured to be 0.88 times that of $P{{H}_{3}}$at the same conditions of temperature and pressure. The gas may be:
(A) ${{C}_{6}}{{H}_{6}}$
(B) $CO$
(C) $N{{O}_{2}}$
(D) ${{N}_{2}}O$

Answer 1

this is based on the rate of diffusion law of Graham and in this first we have to find the mass of the unknown gas X by using the formula: $\dfrac{{{r}_{1}}}{{{r}_{2}}}=\sqrt{\dfrac{{{m}_{2}}}{{{m}_{1}}}}$ and after we get to known about the mass of gas , we can easily identify the gas by calculating the masses of all gases given in the statement. Now solve it.

Complete step by step solution:
It is based on the graham’s law of diffusion. This law states that the rate of diffusion of gases i.e. the time at which the gases diffuse out is directly proportional to the square root of the densities of the gases and inversely proportional to the square root of the masses of the gases. Suppose A and B are the gases having the rate of diffusion as ${{r}_{1}}$ and ${{r}_{2}}$ and densities as ${{d}_{1}}$ and ${{d}_{2}}$and masses as ${{m}_{1}}$ and ${{m}_{2}}$, then;
$\dfrac{{{r}_{1}}}{{{r}_{2}}}=\sqrt{\dfrac{{{d}_{1}}}{{{d}_{2}}}}$ and $\dfrac{{{r}_{1}}}{{{r}_{2}}}=\sqrt{\dfrac{{{m}_{2}}}{{{m}_{1}}}}$
Now considering the numerical,
The rate of diffusion of gas X is 0.888 times the rate of $P{{H}_{3}}$
Suppose x is the rate of diffusion of gas $P{{H}_{3}}$i.e. ${{r}_{P{{H}_{3}}}}$
And rate of diffusion of gas X i.e. ${{r}_{X}}$ will be 0.888x
Mass of the $P{{H}_{3}}$ $= 31+3$
$= 34$
And mass of gas X= y
So, we will apply the formula as;
$\dfrac{{{r}_{1}}}{{{r}_{2}}}=\sqrt{\dfrac{{{m}_{2}}}{{{m}_{1}}}}$
Put all the values in this, we get;
$\dfrac{0.888x}{x}=\sqrt{\dfrac{34}{y}}$
$\dfrac{0.888\times {x}}{{x}}=\sqrt{\dfrac{34}{y}}$
Taking square both sides, we get;
${{(0.888)}^{2}}=\dfrac{34}{y}$
$y=\dfrac{34}{0.7744}$
$= 43.91$
$= 44$
So, the mass of the gas whose rate of diffusion is the same as that of $P{{H}_{3}}$is 44.
Now, we calculate the mass of all the gases, and we can find the gas having this mass.
1. ${{C}_{6}}{{H}_{6}}$ $= 6(12)+ 6$
$= 72+6$
$= 78$
So, it is not that gas.
2. $CO$ $= 12+16$
$= 28$
So, it is not that gas.
3. $N{{O}_{2}}$ $= 14+2(16)$
$= 14+ 32$
$= 46$
So, it is not that gas.
4. ${{N}_{2}}O$ $= 2(14)+16$
$= 28+16$
$= 44$
Since, this gas has that mass no as 44, so ${{N}_{2}}O$ is the gas which has the rate of diffusion 0.88 times that of $P{{H}_{3}}$.

Hence, option(D) is correct.

Note: This law tells us about the relationship between the rate of effusion and diffusion of a gas. By Diffusion we mean the spreading of a gas throughout a volume and by the term effusion we mean the movement of a gas through a tiny hole into an open chamber.