Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

\(2.8 \mathrm{~g}\) of a gas at \(1 \mathrm{~atm}\) and \(273 \mathrm{~K}\) occupies a volume of \(2.24\) litres, the gas can not be: (a) \(\mathrm{O}_{2}\) (b) \(\mathrm{CO}\) (c) \(\mathrm{N}_{2}\) (d) \(\mathrm{C}_{2} \mathrm{H}_{4}\)

Short Answer

Expert verified
Substituting the values into the Ideal Gas Law and solving for molar mass, we find that the gas cannot be (d) C2H4, as its molar mass does not match the calculated molar mass.

Step by step solution

01

Use the Ideal Gas Law

Determine which gas does not fit the observed data by calculating the molar mass. The Ideal Gas Law, which is PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is the temperature, will be used. Given P = 1 atm, V = 2.24 liters, and T = 273 K, we'll first calculate the number of moles (n).
02

Calculate Moles of Gas

Using the Ideal Gas Law, rearrange to calculate n (the number of moles): n = PV/(RT). Substitute in the constants P = 1 atm, V = 2.24 L, R = 0.0821 L atm / (mol K), and T = 273 K to find n.
03

Calculate the Molar Mass of the Gas

Molar mass (M) is calculated by dividing the mass of the gas by the number of moles. Using the formula M = mass/n, where mass = 2.8 g, you can find the molar mass of the unknown gas.
04

Compare Calculated Molar Mass with Given Options

Compare the calculated molar mass with the molar masses of the options provided: (a) O2 = 32 g/mol, (b) CO = 28 g/mol, (c) N2 = 28 g/mol, (d) C2H4 = 28 g/mol + 4 g/mol = 28 + 4(1) = 32 g/mol. The option whose molar mass does not match the calculated molar mass is the gas that the sample cannot be.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molar Mass Calculation
Molar mass is a fundamental concept in chemistry that represents the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of all the atoms in a molecule.

For example, in the calculation of the molar mass of carbon dioxide (CO2), we consider the molar masses of carbon (C) and oxygen (O). Since the atomic mass of carbon is approximately 12.01 g/mol and oxygen is approximately 16.00 g/mol, we calculate it as follows:
M(CO2) = 1(12.01 g/mol) + 2(16.00 g/mol) = 44.01 g/mol.

To find the molar mass of a gas, as reflected in the step by step solution, we divide the given mass of the gas by the number of moles computed using the ideal gas law equation. This gives us a specific value that can be compared with known molar masses to identify the gas.
Gas Constant (R)
The gas constant, symbolized as R, is a crucial figure in the ideal gas law. It's the universal proportionality constant that relates the energy scale in physics to the temperature scale. Its value depends on the units used for pressure, volume, and temperature. In the context of the ideal gas law, R is often expressed in liters atmospheres per mole kelvin (L atm / (mol K)).

For the ideal gas law to be applied accurately, you must use the correct value of R that corresponds with your units for pressure, volume, and temperature. The commonly used value is 0.0821 L atm / (mol K), which is used when pressure is measured in atmospheres, volume in liters, and temperature in Kelvin.

Utilizing R allows us to derive the number of moles in a given sample of gas when we have measurements of pressure, volume, and temperature, thus bridging these gas properties with the mole concept and molar mass calculation.
Mole Concept
The mole concept is a bridge between the microscopic world of atoms and molecules and the macroscopic world we observe. One mole is defined as the amount of substance that contains as many entities (atoms, molecules, or other particles) as there are atoms in 12 grams of carbon-12 (6.022 x 1023 entities, known as Avogadro's number).

The ability to convert between moles and mass is a core skill in chemistry. This conversion is straightforward when the molar mass of the substance is known. To find the moles from mass, you divide the mass by the molar mass. Conversely, to find the mass from moles, you multiply the moles by the molar mass. In the context of the textbook problem provided, understanding the mole concept is essential for using the Ideal Gas Law equation to calculate the molar mass and identify the unknown gas.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A certain sample of gas has a volume of \(0.2\) litre measured at 1 atm pressure and \(0^{\circ} \mathrm{C}\). At the same pressure but at \(273^{\circ} \mathrm{C}\), its volume will be: (a) \(0.4\) litre (b) \(0.8\) litre (c) \(27.8\) litres (d) \(55.6\) litres

A \(0.50 \mathrm{~L}\) container is occupied by nitrogen at a pressure of 800 torr and a temperature of \(0^{\circ} \mathrm{C}\). The container can only withstand a pressure of \(3.0 \mathrm{~atm} .\) What is the highest temperature \(\left({ }^{\circ} \mathrm{C}\right)\) to which the container may be heated? (a) 505 (b) 450 (c) 625 (d) 560

The density of gas \(A\) is twice that to \(B\) at the same temperature the molecular weight of gas \(B\) is twice that of \(A\). The ratio of pressure of gas \(A\) and \(B\) will be : (a) \(1: 6\) (b) \(1: 1\) (c) \(4: 1\) (d) \(1: 4\)

A gaseous mixture contains 4 molecules with a velocity of \(6 \mathrm{~cm} \mathrm{sec}^{-1}, 5\) molecules with a velocity of \(2 \mathrm{~cm} \mathrm{sec}^{-1}\) and 10 molecules with a velocity of \(3 \mathrm{~cm} \mathrm{sec}^{-1}\). What is the RMS velocity of the gas: (a) \(2.5 \mathrm{~cm} \mathrm{sec}^{-1}\) (b) \(1.9 \mathrm{~cm} \mathrm{sec}^{-1}\) (c) \(3.6 \mathrm{~cm} \mathrm{sec}^{-1}\) (d) \(4.6 \mathrm{~cm} \mathrm{sec}^{-1}\)

The critical temperature of a substance is: (a) The temperature above which the substance undergoes decomposition (b) The temperature above which a substance can exist only in gaseous state (c) Boiling point of the substance (d) All are wrong

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free