If the rate of diffusion of \[A\]is $5$ times that of $B$, what will be the density ratio of $A$ and $B$.
A.$\dfrac{1}{{25}}$
B.$\dfrac{1}{5}$
C.$25$
D.$5$
Answer
515.7k+ views
Hint: The relation between its density and the rate of diffusion of a gas can be represented by Graham's law which states that "The rate of diffusion of a gas is inversely proportional to the square root of its density under given conditions of temperature and pressure.
Complete answer:
Graham's law states that the rate of diffusion or of effusion of a gas is inversely proportional to the square root of its molecular weight. Thus, if the molecular weight of one gas is four times that of another, it would diffuse through a porous plug or escape through a small pinhole in a vessel at half the rate of the other (heavier gases diffuse more slowly).
In the same conditions of temperature and pressure, the molar mass is proportional to the mass density. Therefore, the rates of diffusion of different gases are inversely proportional to the square roots of their mass densities.
$r\alpha \dfrac{1}{{\sqrt d }}{r_A} = 5{r_B}$
The rate of diffusion of gas is directly proportional to the square root of its density.
$
\Rightarrow \dfrac{{{r_A}}}{{{r_B}}} = \sqrt {\dfrac{{{d_B}}}{{{d_A}}}} \\
\Rightarrow \dfrac{{5{r_B}}}{{{r_B}}} = \sqrt {\dfrac{{{d_B}}}{{{d_A}}}} \\
\Rightarrow \dfrac{{{d_B}}}{{{d_A}}} = \dfrac{{25}}{1} \\
$
Or, $\dfrac{{{d_A}}}{{{d_B}}} = \dfrac{1}{{25}}$
So, the correct answer is (A)$\dfrac{1}{{25}}$
Note:
If the medium that a given particle has to diffuse through is very dense or viscous, then the particle will have a harder time diffusing through it. So the rate of diffusion will be lower. If the medium is less dense or less viscous, then the particles will be able to move more quickly and will diffuse faster.
Complete answer:
Graham's law states that the rate of diffusion or of effusion of a gas is inversely proportional to the square root of its molecular weight. Thus, if the molecular weight of one gas is four times that of another, it would diffuse through a porous plug or escape through a small pinhole in a vessel at half the rate of the other (heavier gases diffuse more slowly).
In the same conditions of temperature and pressure, the molar mass is proportional to the mass density. Therefore, the rates of diffusion of different gases are inversely proportional to the square roots of their mass densities.
$r\alpha \dfrac{1}{{\sqrt d }}{r_A} = 5{r_B}$
The rate of diffusion of gas is directly proportional to the square root of its density.
$
\Rightarrow \dfrac{{{r_A}}}{{{r_B}}} = \sqrt {\dfrac{{{d_B}}}{{{d_A}}}} \\
\Rightarrow \dfrac{{5{r_B}}}{{{r_B}}} = \sqrt {\dfrac{{{d_B}}}{{{d_A}}}} \\
\Rightarrow \dfrac{{{d_B}}}{{{d_A}}} = \dfrac{{25}}{1} \\
$
Or, $\dfrac{{{d_A}}}{{{d_B}}} = \dfrac{1}{{25}}$
So, the correct answer is (A)$\dfrac{1}{{25}}$
Note:
If the medium that a given particle has to diffuse through is very dense or viscous, then the particle will have a harder time diffusing through it. So the rate of diffusion will be lower. If the medium is less dense or less viscous, then the particles will be able to move more quickly and will diffuse faster.
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