
At NTP, sample of chlorine and oxygen is taken. Given that they both were of equal volumes, what will be the ratio of their number of molecules?
A. $1:1$
B. $32:27$
C. $2:1$
D. $16:14$
Answer
163.8k+ views
Hint:According to ideal gas law, $PV = nRT$ , where $P$ is the pressure of the gas, $V$ is its volume, $n$ is the number of moles, $T$ is temperature and $R$ is the universal gas constant having a value of $8.314{\text{ J}}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}$ . Use this law to calculate the required ratio.
Formula used:
Ideal gas law,
$PV = nRT$
Here $P$ is the pressure of the gas,
$V$ is the volume of the gas,
$n$ is the number of moles,
$T$ is the temperature and
$R$ is the universal gas constant having a value of $8.314{\text{ J}}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}$ .
Complete answer:
According to ideal gas law,
$PV = nRT$ … (1)
It is given that a sample of equal volume of chlorine and oxygen is taken.
Let this volume be $V$ .
It is also given that the sample is taken at NTP, the normal temperature and pressure.
Thus, both the temperature and pressure of the two gases are the same, that is, $293.15{\text{ K}}$ and $1{\text{ atm}}$ .
Let these values be $T$ and $P$ respectively.
Now, let the number of molecules of chlorine and oxygen be ${n_1}$ and ${n_2}$ respectively.
Then, using ideal gas law, given in formula (1),
Number of molecules of chlorine, ${n_1} = \dfrac{{PV}}{{RT}}$ … (2)
Similarly,
Number of molecules of oxygen, ${n_2} = \dfrac{{PV}}{{RT}}$ … (3)
Dividing (2) by (3), we get:
\[\dfrac{{{n_1}}}{{{n_2}}} = 1\]
Hence, the ratio of the number of moles and hence molecules of chlorine and oxygen is 1:1.
Thus, the correct option is A.
Note: According to Avogadro’s Postulate, all equal volumes of gases at the same temperature and pressure contain the same number of molecules or moles. You could have used this postulate to solve the above question as well.
Formula used:
Ideal gas law,
$PV = nRT$
Here $P$ is the pressure of the gas,
$V$ is the volume of the gas,
$n$ is the number of moles,
$T$ is the temperature and
$R$ is the universal gas constant having a value of $8.314{\text{ J}}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}$ .
Complete answer:
According to ideal gas law,
$PV = nRT$ … (1)
It is given that a sample of equal volume of chlorine and oxygen is taken.
Let this volume be $V$ .
It is also given that the sample is taken at NTP, the normal temperature and pressure.
Thus, both the temperature and pressure of the two gases are the same, that is, $293.15{\text{ K}}$ and $1{\text{ atm}}$ .
Let these values be $T$ and $P$ respectively.
Now, let the number of molecules of chlorine and oxygen be ${n_1}$ and ${n_2}$ respectively.
Then, using ideal gas law, given in formula (1),
Number of molecules of chlorine, ${n_1} = \dfrac{{PV}}{{RT}}$ … (2)
Similarly,
Number of molecules of oxygen, ${n_2} = \dfrac{{PV}}{{RT}}$ … (3)
Dividing (2) by (3), we get:
\[\dfrac{{{n_1}}}{{{n_2}}} = 1\]
Hence, the ratio of the number of moles and hence molecules of chlorine and oxygen is 1:1.
Thus, the correct option is A.
Note: According to Avogadro’s Postulate, all equal volumes of gases at the same temperature and pressure contain the same number of molecules or moles. You could have used this postulate to solve the above question as well.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Charging and Discharging of Capacitor

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

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
