Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 82

A sample of \(\mathrm{N}_{2}(\mathrm{g})\) effuses through a tiny hole in \(38 \mathrm{s}\) What must be the molar mass of a gas that requires \(64 \mathrm{s}\) to effuse under identical conditions?

Short Answer

Expert verified
The molar mass of the unknown gas, that effuses in 64 seconds under the same conditions as nitrogen, is about 47.3 g/mol.

Step by step solution

01

Understanding Graham's law of effusion

Graham’s law of effusion states that, under the same conditions of temperature and pressure, the rate of effusion of different gases is inversely proportional to the square root of their molar masses. Mathematically this can be written as: \( r_1 / r_2 = \sqrt {M_2 / M_1} \) where r is the rate of effusion and M is the molar mass.
02

Apply the rates and molar mass given

Here, the rate of effusion is proportional to 1/time required. So we can express \( r_1 = 1/t_1 \) and \( r_2 = 1/t_2 \). Given in the problem, \( r_1 = 1/38s \) and \( r_2 = 1/64s \). The first gas is nitrogen with a molar mass \( M_1 = 28 g/mol \). Substituting these into Graham’s law equation we get: \( (1/38s) / (1/64s) = \sqrt {M_2 / 28 g/mol} \)
03

Solve the equation

Rearranging to solve for \( M_2 \), we get: \( M_2 = [(64/38)^2] * 28 g/mol \). Solving this, we get \( M_2 = 47.3 g/mol \). So the molar mass of the unknown gas is about 47.3 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

Explain why it is necessary to include the density of \(\mathrm{Hg}(1)\) and the value of the acceleration due to gravity, \(g,\) in a precise definition of a millimeter of mercury (page 194 ).

A particular gaseous hydrocarbon that is \(82.7 \%\) C and \(17.3 \%\) H by mass has a density of \(2.33 \mathrm{g} / \mathrm{L}\) at \(23^{\circ} \mathrm{C}\) and \(746 \mathrm{mm} \mathrm{Hg} .\) What is the molecular formula of this hydrocarbon?

What is the volume, in milliliters, occupied by \(89.2 \mathrm{g} \mathrm{CO}_{2}(\mathrm{g})\) at \(37^{\circ} \mathrm{C}\) and \(737 \mathrm{mmHg} ?\)

You purchase a bag of potato chips at an ocean beach to take on a picnic in the mountains. At the picnic, you notice that the bag has become inflated, almost to the point of bursting. Use your knowledge of gas behavior to explain this phenomenon.

A constant-volume vessel contains \(12.5 \mathrm{g}\) of a gas at 21 ^ C. If the pressure of the gas is to remain constant as the temperature is raised to \(210^{\circ} \mathrm{C}\), how many grams of gas must be released?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Physical Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Kinetics

Read Explanation

Making Measurements

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