
The ratio of rate of diffusion of gases A and B is ${\text{1:4}}$. If the ratio of their masses present in the mixture is ${\text{2:3}}$, what is the ratio of their mole fraction?
A) $\dfrac{{\text{1}}}{{\text{8}}}$
B) $\dfrac{{\text{1}}}{{{\text{12}}}}$
C) $\dfrac{{\text{1}}}{{{\text{16}}}}$
D) $\dfrac{{\text{1}}}{{{\text{24}}}}$
Answer
578.1k+ views
Hint: The change in number of molecules that diffuse per unit time is known as the rate of diffusion. We can apply Graham's law which states that the rate of diffusion of any gas is inversely proportional to the square root of the molar mass of gas.
Formula used: $\dfrac{{{R_1}}}{{{R_2}}} = \sqrt {\dfrac{{{M_2}}}{{{M_1}}}} $
Complete step by step answer:
Graham’s law states that the rate of diffusion of any gas is inversely proportional to the square root of the molar mass of gas. Thus,
$\dfrac{{{R_1}}}{{{R_2}}} = \sqrt {\dfrac{{{M_2}}}{{{M_1}}}} $
Where, ${R_1}$ and ${R_2}$ are the rates of diffusion,
${M_1}$ and ${M_2}$ are the molar masses.
The ratio of rate of diffusion of gases A and B is ${\text{1:4}}$. Thus,
$\dfrac{1}{4} = \sqrt {\dfrac{{{M_2}}}{{{M_1}}}} $
$\dfrac{1}{{16}} = \dfrac{{{M_2}}}{{{M_1}}}$
$\dfrac{{{M_1}}}{{{M_2}}} = \dfrac{{16}}{1}$ …… (1)
The ratio of molar masses of the gases present in the mixture is ${\text{2:3}}$. Thus,
$\dfrac{{{W_1}}}{{{W_2}}} = \dfrac{2}{3}$ …… (2)
The number of moles is the ratio of mass to the molar mass.
Thus, we can obtain the number of moles of gases by dividing equation (2) with equation (1). Thus,
$\dfrac{{{n_1}}}{{{n_2}}} = \dfrac{{\dfrac{{{W_1}}}{{{W_2}}}}}{{\dfrac{{{M_1}}}{{{M_2}}}}}$
$\dfrac{{{n_1}}}{{{n_2}}} = \dfrac{{\dfrac{2}{3}}}{{\dfrac{{16}}{1}}}$
$\dfrac{{{n_1}}}{{{n_2}}} = \dfrac{2}{3} \times \dfrac{1}{{16}}$
$\dfrac{{{n_1}}}{{{n_2}}} = \dfrac{1}{{24}}$
Thus, the ratio of their mole fraction is $\dfrac{1}{{24}}$.
Thus, the correct option is (D) $\dfrac{1}{{24}}$.
Note: Rate of diffusion varies with different factors. Various factors that affect the rate of diffusion are as follows:
- Temperature: As the temperature increases, the movement of molecules increases. Thus, the rate of diffusion increases. Thus, the rate of diffusion is directly proportional to the temperature.
- Density of solvent: As the density of the solvent increases, the movement of molecules decreases. Thus, the rate of diffusion decreases. Thus, the rate of diffusion is inversely proportional to the density of solvent.
Formula used: $\dfrac{{{R_1}}}{{{R_2}}} = \sqrt {\dfrac{{{M_2}}}{{{M_1}}}} $
Complete step by step answer:
Graham’s law states that the rate of diffusion of any gas is inversely proportional to the square root of the molar mass of gas. Thus,
$\dfrac{{{R_1}}}{{{R_2}}} = \sqrt {\dfrac{{{M_2}}}{{{M_1}}}} $
Where, ${R_1}$ and ${R_2}$ are the rates of diffusion,
${M_1}$ and ${M_2}$ are the molar masses.
The ratio of rate of diffusion of gases A and B is ${\text{1:4}}$. Thus,
$\dfrac{1}{4} = \sqrt {\dfrac{{{M_2}}}{{{M_1}}}} $
$\dfrac{1}{{16}} = \dfrac{{{M_2}}}{{{M_1}}}$
$\dfrac{{{M_1}}}{{{M_2}}} = \dfrac{{16}}{1}$ …… (1)
The ratio of molar masses of the gases present in the mixture is ${\text{2:3}}$. Thus,
$\dfrac{{{W_1}}}{{{W_2}}} = \dfrac{2}{3}$ …… (2)
The number of moles is the ratio of mass to the molar mass.
Thus, we can obtain the number of moles of gases by dividing equation (2) with equation (1). Thus,
$\dfrac{{{n_1}}}{{{n_2}}} = \dfrac{{\dfrac{{{W_1}}}{{{W_2}}}}}{{\dfrac{{{M_1}}}{{{M_2}}}}}$
$\dfrac{{{n_1}}}{{{n_2}}} = \dfrac{{\dfrac{2}{3}}}{{\dfrac{{16}}{1}}}$
$\dfrac{{{n_1}}}{{{n_2}}} = \dfrac{2}{3} \times \dfrac{1}{{16}}$
$\dfrac{{{n_1}}}{{{n_2}}} = \dfrac{1}{{24}}$
Thus, the ratio of their mole fraction is $\dfrac{1}{{24}}$.
Thus, the correct option is (D) $\dfrac{1}{{24}}$.
Note: Rate of diffusion varies with different factors. Various factors that affect the rate of diffusion are as follows:
- Temperature: As the temperature increases, the movement of molecules increases. Thus, the rate of diffusion increases. Thus, the rate of diffusion is directly proportional to the temperature.
- Density of solvent: As the density of the solvent increases, the movement of molecules decreases. Thus, the rate of diffusion decreases. Thus, the rate of diffusion is inversely proportional to the density of solvent.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

10 examples of friction in our daily life

