Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 43

At STP, \(0.280 \mathrm{~L}\) of a gas weighs \(0.400 \mathrm{~g}\). Calculate the molar mass of the gas.

Short Answer

Expert verified
The molar mass of the gas is \(32 \mathrm{~g/mol}\).

Step by step solution

01

Understanding the problem

We have the volume (0.280 L) and the weight (0.400 g) of the gas at STP. We know that molar volume of any ideal gas at STP is \(22.4 \mathrm{~L/mol}\). The molar mass is defined as the mass of one mole of a substance and it is measured in g/mol. So, we need to find out how many moles are there in 0.280 L and then calculate the mass of that amount of gas in grams. This will yield us the molar mass.
02

Calculate the moles of the gas

We have the volume of the gas (0.280 L) and we know that molar volume of any ideal gas at STP is \(22.4 \mathrm{~L/mol}\). Therefore, the number of moles(n) of this gas can be calculated as follows: \(n = \frac{Volume~of ~gas}{Molar~volume~at~STP} = \frac{0.280~L}{22.4~L/mol} = 0.0125~mol\)
03

Calculate the molar mass

Molar mass is the mass of 1 mole of a substance. We know the mass of 0.0125 moles of this gas (which is 0.400 g). Therefore, the molar mass(M) of this gas can be calculated as follows: \(M = \frac{Mass~of ~gas}{number~of~moles} = \frac{0.400~g}{0.0125~mol} = 32~g/mol\). Thus, the molar mass of the gas is \(32 \mathrm{~g/mol}\).

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.

Start your free trial

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

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

One way to gain a physical understanding of \(b\) in the van der Waals equation is to calculate the "excluded volume." Assume that the distance of closest approach between two similar atoms is the sum of their radii \((2 r) .\) (a) Calculate the volume around each atom into which the center of another atom cannot penetrate. (b) From your result in (a), calculate the excluded volume for 1 mole of the atoms, which is the constant \(b\). How does this volume compare with the sum of the volumes of 1 mole of the atoms?

Under the same conditions of temperature and pressure, why does \(1 \mathrm{~L}\) of moist air weigh less than \(1 \mathrm{~L}\) of dry air? In weather forecasts, an oncoming lowpressure front usually means imminent rainfall. Explain.

Nitrous oxide \(\left(\mathrm{N}_{2} \mathrm{O}\right)\) can be obtained by the thermal decomposition of ammonium nitrate \(\left(\mathrm{NH}_{4} \mathrm{NO}_{3}\right) .\) (a) Write a balanced equation for the reaction. (b) In a certain experiment, a student obtains \(0.340 \mathrm{~L}\) of the gas at \(718 \mathrm{mmHg}\) and \(24^{\circ} \mathrm{C}\). If the gas weighs \(0.580 \mathrm{~g},\) calculate the value of the gas constant.

A quantity of \(0.225 \mathrm{~g}\) of a metal \(\mathrm{M}\) (molar mass = \(27.0 \mathrm{~g} / \mathrm{mol}\) ) liberated \(0.303 \mathrm{~L}\) of molecular hydrogen (measured at \(17^{\circ} \mathrm{C}\) and \(741 \mathrm{mmHg}\) ) from an excess of hydrochloric acid. Deduce from these data the corresponding equation and write formulas for the oxide and sulfate of M.

What is the difference between gas diffusion and effusion? State Graham's law and define the terms in Equation (5.17)
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Chemical Analysis

Read Explanation

Kinetics

Read Explanation

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Inorganic Chemistry

Read Explanation
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