
Under identical conditions of pressure and temperature, 4L of gaseous mixture ( ${H_2}$ and $C{H_4}$ ) effuses through a hole in 5 min whereas 4L of a gas X of molecular mass 36 takes 10 min to effuse through the same hole. The mole ratio of ${H_2}:C{H_4}$ in the mixture is:
A. $1:3$
B. $1:2$
C. $2:3$
D. $1:1$
Answer
233.1k+ views
Hint: Basically, effusion is the ability of a gas to escape or travel through a small hole with a small aperture and from a place of high concentration to low concentration. Generally, this is a process that happens every day in our lives.
Formula used:
Time $\alpha \sqrt {molecular\,weight} $
Molecular weight ${m_1} = \dfrac{{{n_1}{M_1} + {n_2}{M_2}}}{{{n_1} + {n_2}}}$
Where ${n_1}$ and ${n_2}$ are the number of moles of ${H_2}$ and $C{H_2}$ respectively.
Complete step by step answer:
Basically effusion in simple words is the ability of gas to travel through a small opening. It occurs when the size of the aperture of hole is smaller than the mean free path of the molecules. It generally occurs or is facilitated by a difference of pressures.
Now, we have to calculate the mole ratio of ${H_2}:C{H_4}$ in the mixture.
So, rate of diffusion is directly proportional to $\dfrac{1}{{\sqrt {molecular\,weight} }}$
Therefore, time $\alpha \sqrt {molecular\,weight} $
Now, we have been given that the time taken for a mixture is $5\min $whereas the time taken for unknown gas is 10min.
So, $ = \sqrt {\dfrac{{36}}{{10}}} $
Which means, $\sqrt {{m_1} = 3} $
Therefore, ${m_1} = 9$
Now, according to the formula
Molecular weight ${m_1} = \dfrac{{{n_1}{M_1} + {n_2}{M_2}}}{{{n_1} + {n_2}}}$
Now, substitute the value,
$9 = \dfrac{{{n_1} \times 2 + {n_2} \times 16}}{{({n_1} + {n_2})}}$
So, $9{n_1} + 9{n_2} = 2{n_1} + 16{n_2}$
$7{n_1} = {n_2}$
Therefore, ${n_1} = {n_2}$
So, the mole ratio is $1:1$
Hence, option D is correct.
Note:
Don’t get confused with diffusion and effusion. These two terms are nothing but two different properties of gases. Diffusion is the ability of gases to mix with each other without requiring bulk motion whereas effusion is the ability of gas to escape or travel through a small hole with a small aperture.
Formula used:
Time $\alpha \sqrt {molecular\,weight} $
Molecular weight ${m_1} = \dfrac{{{n_1}{M_1} + {n_2}{M_2}}}{{{n_1} + {n_2}}}$
Where ${n_1}$ and ${n_2}$ are the number of moles of ${H_2}$ and $C{H_2}$ respectively.
Complete step by step answer:
Basically effusion in simple words is the ability of gas to travel through a small opening. It occurs when the size of the aperture of hole is smaller than the mean free path of the molecules. It generally occurs or is facilitated by a difference of pressures.
Now, we have to calculate the mole ratio of ${H_2}:C{H_4}$ in the mixture.
So, rate of diffusion is directly proportional to $\dfrac{1}{{\sqrt {molecular\,weight} }}$
Therefore, time $\alpha \sqrt {molecular\,weight} $
Now, we have been given that the time taken for a mixture is $5\min $whereas the time taken for unknown gas is 10min.
So, $ = \sqrt {\dfrac{{36}}{{10}}} $
Which means, $\sqrt {{m_1} = 3} $
Therefore, ${m_1} = 9$
Now, according to the formula
Molecular weight ${m_1} = \dfrac{{{n_1}{M_1} + {n_2}{M_2}}}{{{n_1} + {n_2}}}$
Now, substitute the value,
$9 = \dfrac{{{n_1} \times 2 + {n_2} \times 16}}{{({n_1} + {n_2})}}$
So, $9{n_1} + 9{n_2} = 2{n_1} + 16{n_2}$
$7{n_1} = {n_2}$
Therefore, ${n_1} = {n_2}$
So, the mole ratio is $1:1$
Hence, option D is correct.
Note:
Don’t get confused with diffusion and effusion. These two terms are nothing but two different properties of gases. Diffusion is the ability of gases to mix with each other without requiring bulk motion whereas effusion is the ability of gas to escape or travel through a small hole with a small aperture.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Hydrocarbons Class 11 Chemistry Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Chemistry Chapter 5 CBSE Notes - 2025-26

Equilibrium Class 11 Chemistry Chapter 6 CBSE Notes - 2025-26

Organic Chemistry Some Basic Principles And Techniques Class 11 Chemistry Chapter 8 CBSE Notes - 2025-26

NCERT Solutions For Class 11 Chemistry Chapter 7 Redox Reactions (2025-26)

