Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 88

A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required \(105 \mathrm{~s}\) for $1.0 \mathrm{~L}$ of the gas to effuse. Under identical experimental conditions it required \(31 \mathrm{~s}\) for \(1.0 \mathrm{~L}\) of \(\mathrm{O}_{2}\) gas to effuse. Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of $1.0 \mathrm{~L} ;$ in other words, rate is the amount that diffuses over the time it takes to diffuse.)

Short Answer

Expert verified
The molar mass of the unknown gas can be calculated using Graham's Law of Effusion. By comparing the rates of effusion of the unknown gas and \(\mathrm{O}_{2}\) gas (\( \frac{1}{105} \frac{L}{s}\) and \( \frac{1}{31} \frac{L}{s}\), respectively), we can set up the equation: \( \frac{ \frac{1}{105} }{ \frac{1}{31} } = \sqrt{\frac{32}{M_{unknown}}} \). Solving for \(M_{unknown}\), we find that the molar mass of the unknown gas is approximately 114.11 g/mol.

Step by step solution

01

Understand Graham's Law of Effusion

Graham's Law of Effusion states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, if we have two gases A and B, the formula can be written as: \( \frac{Rate_{A}}{Rate_{B}} = \sqrt{\frac{M_{B}}{M_{A}}} \) Where \(Rate_{A}\) and \(Rate_{B}\) are the rates of effusion of gases A and B, respectively, and \(M_{A}\) and \(M_{B}\) are their corresponding molar masses.
02

Calculate the rates of effusion for both gases

We are given the times required for 1.0 L of both the unknown gas and \(\mathrm{O}_{2}\) gas to effuse. We can find the rates of effusion by taking the reciprocal of the given times: Rate of effusion for unknown gas = \( \frac{1}{105} \frac{L}{s} \) Rate of effusion for \(\mathrm{O}_{2}\) gas = \( \frac{1}{31} \frac{L}{s} \)
03

Set up the equation using Graham's Law of Effusion

Now that we have the rates of effusion for both gases, we can set up the equation using Graham's Law: \( \frac{ \frac{1}{105} }{ \frac{1}{31} } = \sqrt{\frac{32}{M_{unknown}}} \) Here, 32 is the molar mass of \(\mathrm{O}_{2}\).
04

Solve for the molar mass of the unknown gas

We will solve for \(M_{unknown}\): \( \frac{1}{105} \times 31 = \sqrt{\frac{32}{M_{unknown}}} \) \( \frac{31}{105} = \sqrt{\frac{32}{M_{unknown}}} \) Square both sides: \( (\frac{31}{105})^2 = \frac{32}{M_{unknown}} \) Now, isolate \(M_{unknown}\) on one side of the equation: \( M_{unknown} = \frac{32}{(\frac{31}{105})^2} \)
05

Calculate the molar mass of the unknown gas

Finally, plug in the numbers and calculate the molar mass of the unknown gas: \( M_{unknown} = \frac{32}{(\frac{31}{105})^2} = 114.11 \) Therefore, the molar mass of the unknown gas is approximately 114.11 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

Cyclopropane, a gas used with oxygen as a general anesthetic, is composed of \(85.7 \% \mathrm{C}\) and \(14.3 \% \mathrm{H}\) by mass. \((\mathbf{a})\) If $1.56 \mathrm{~g}\( of cyclopropane has a volume of \)1.00 \mathrm{~L}$ at 99.7 \(\mathrm{kPa}\) and \(50.0^{\circ} \mathrm{C}\), what is the molecular formula of cyclopropane? (b) Judging from its molecular formula, would you expect cyclopropane to deviate more or less than Ar from ideal-gas behavior at moderately high pressures and room temperature? Explain. (c) Would cyclopropane effuse through a pinhole faster or more slowly than methane, \(\mathrm{CH}_{4} ?\)

A deep-sea diver uses a gas cylinder with a volume of \(10.0 \mathrm{~L}\) and a content of \(51.2 \mathrm{~g}\) of \(\mathrm{O}_{2}\) and \(32.6 \mathrm{~g}\) of He. Calculate the partial pressure of each gas and the total pressure if the temperature of the gas is \(19^{\circ} \mathrm{C}\).

(a) What are the mole fractions of each component in a mixture of $15.08 \mathrm{~g}\( of \)\mathrm{O}_{2}, 8.17 \mathrm{~g}\( of \)\mathrm{N}_{2},$ and \(2.64 \mathrm{~g}\) of \(\mathrm{H}_{2}\) (b) What is the partial pressure in atm of each component of this mixture if it is held in a \(15.50-\mathrm{L}\) vessel at \(15^{\circ} \mathrm{C} ?\)

Which statement concerning the van der Waals constants \(a\) and \(b\) is true? (a) The magnitude of \(a\) relates to molecular volume, whereas \(b\) relates to attractions between molecules. (b) The magnitude of \(a\) relates to attractions between molecules, whereas \(b\) relates to molecular volume. (c) The magnitudes of \(a\) and \(b\) depend on pressure. (d) The magnitudes of \(a\) and \(b\) depend on temperature.

A sample of \(5.00 \mathrm{~mL}\) of diethylether $\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OC}_{2} \mathrm{H}_{5},\right.\( density \)=0.7134 \mathrm{~g} / \mathrm{mL}\( ) is introduced into a \)6.00-\mathrm{L}$ vessel that already contains a mixture of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\), whose partial pressures are \(R_{\mathrm{N}_{2}}=21.08 \mathrm{kPa}\) and \(P_{\mathrm{O}_{2}}=76.1 \mathrm{kPa}\). The temperature is held at \(35.0^{\circ} \mathrm{C},\) and the diethylether totally evaporates. (a) Calculate the partial pressure of the diethylether. (b) Calculate the total pressure in the container.
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Physical Chemistry

Read Explanation

Chemical Analysis

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