
0.56 g of gas occupies \[{\rm{280c}}{{\rm{m}}^{\rm{3}}}\] at NTP. Then its molecular mass is:
A. 4.8
B. 44.8
C. 11.2
D. 22.4
Answer
163.5k+ views
Hint: The NTP denotes normal temperature and pressure. One mole of an ideal gas has a molar volume at NTP is 22.4 L. The temperature and pressure at NTP are 273 K and 1 atm respectively.
Complete Step by Step Solution:
Here in this question, we are given an amount of gas that occupies a certain amount of volume at NTP.
We have to find out the molecular mass of the gas.
We will have to convert \[{\rm{280c}}{{\rm{m}}^{\rm{3}}}\] into litres.
\[1c{m^3}\] is equal to 0.001 L.
So, \[{\rm{280c}}{{\rm{m}}^{\rm{3}}}\] is equal to 0.28 L.
We know that 1 mole of a gas at NTP occupies 22.4 L of volume.
We also know that one mole of a substance is its molecular mass.
0.28 L occupies 0.56 g of the ideal gas.
1 L will occupy \[\frac{{0.56g}}{{0.28L}}\] of the ideal gas.
So, 22.4 L will occupy \[\frac{{0.56g}}{{0.28L}}(22.4L)\] of the ideal gas.
Hence, 22.4 L will occupy 44.8 g of the ideal gas.
Therefore, the molecular mass of the ideal gas is 44.8g.
So, option B is correct.
Note: STP means standard temperature and pressure. At standard temperature and pressure, a substance carries a temperature of 273 K and the pressure is 1 atmosphere. This definition is given by The International Union of Pure and Applied Chemistry (IUPAC). The National Institute of Standards and Technology, however, specifies STP as 1 atm and \[20^\circ C\] (293.15 K, or \[68^\circ F\]).
Complete Step by Step Solution:
Here in this question, we are given an amount of gas that occupies a certain amount of volume at NTP.
We have to find out the molecular mass of the gas.
We will have to convert \[{\rm{280c}}{{\rm{m}}^{\rm{3}}}\] into litres.
\[1c{m^3}\] is equal to 0.001 L.
So, \[{\rm{280c}}{{\rm{m}}^{\rm{3}}}\] is equal to 0.28 L.
We know that 1 mole of a gas at NTP occupies 22.4 L of volume.
We also know that one mole of a substance is its molecular mass.
0.28 L occupies 0.56 g of the ideal gas.
1 L will occupy \[\frac{{0.56g}}{{0.28L}}\] of the ideal gas.
So, 22.4 L will occupy \[\frac{{0.56g}}{{0.28L}}(22.4L)\] of the ideal gas.
Hence, 22.4 L will occupy 44.8 g of the ideal gas.
Therefore, the molecular mass of the ideal gas is 44.8g.
So, option B is correct.
Note: STP means standard temperature and pressure. At standard temperature and pressure, a substance carries a temperature of 273 K and the pressure is 1 atmosphere. This definition is given by The International Union of Pure and Applied Chemistry (IUPAC). The National Institute of Standards and Technology, however, specifies STP as 1 atm and \[20^\circ C\] (293.15 K, or \[68^\circ F\]).
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

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Types of Solutions

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

NCERT Solutions for Class 11 Chemistry In Hindi Chapter 1 Some Basic Concepts of Chemistry

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

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Degree of Dissociation and Its Formula With Solved Example for JEE

NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reaction
