/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 105 The vapor pressure of a volatile... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 105

The vapor pressure of a volatile liquid can be determined by slowly bubbling a known volume of gas through it at a known temperature and pressure. In an experiment, \(5.00 \mathrm{~L}\) of \(\mathrm{N}_{2}\) gas is passed through \(7.2146 \mathrm{~g}\) of liquid benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}\), at \(26.0{ }^{\circ} \mathrm{C}\). The liquid remaining after the experiment weighs \(5.1493 \mathrm{~g}\). Assuming that the gas becomes saturated with benzene vapor and that the total gas volume and temperature remain constant, what is the vapor pressure of the benzene in torr?

Short Answer

Expert verified
The vapor pressure of benzene is approximately \(98.49 \, \mathrm{torr}\).

Step by step solution

01

Calculate the mass of evaporated benzene

Subtract the final mass of liquid benzene from the initial mass to find the mass of benzene that has evaporated: Mass of evaporated benzene = Initial mass - Final mass Mass of evaporated benzene = \(7.2146 \mathrm{~g}\) - \(5.1493 \mathrm{~g}\) = \(2.0653 \mathrm{~g}\)
02

Determine the amount of benzene in moles

To calculate the amount of benzene in moles, use the molar mass of benzene, which is approximately \(78.11 \mathrm{~g/mol}\). Moles of benzene = Mass of evaporated benzene / Molar mass of benzene Moles of benzene = \(2.0653 \mathrm{~g}\) / \(78.11 \mathrm{~g/mol}\) = \(0.026451 \mathrm{~mol}\)
03

Apply the ideal gas law

Use the ideal gas law formula, PV = nRT, to find the pressure of the benzene vapor. The values for R (the ideal gas constant) and T (temperature in Kelvin) are given. Convert the temperature from Celsius to Kelvin first: Temperature in Kelvin = \(26.0{ }^{\circ} \mathrm{C} + 273.15 = 299.15 \, \mathrm{K}\) Now, rearrange the ideal gas law equation to solve for pressure, P: P = nRT / V
04

Calculate the vapor pressure

Use the values of moles of benzene (n), temperature (T), volume of nitrogen gas (V), and the ideal gas constant (R) to find the vapor pressure of benzene: P = \((0.026451 \, \mathrm{mol}) (0.0821 \, \mathrm{L \cdot atm/mol \cdot K})(299.15 \, \mathrm{K}) / (5.00 \, \mathrm{L}) = 0.1296 \, \mathrm{atm}\)
05

Convert the pressure to torr

In order to convert the vapor pressure of benzene from atm to torr, use the conversion factor \(1 \, \mathrm{atm} = 760 \, \mathrm{torr}\): Vapor pressure of benzene = \((0.1296 \, \mathrm{atm}) (760 \, \mathrm{torr/atm}) = 98.49 \, \mathrm{torr}\) The vapor pressure of benzene is approximately \(98.49 \, \mathrm{torr}\).

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 91影视!

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

Butane and 2 -methylpropane, whose space-filling models are shown, are both nonpolar and have the same molecular formula, yet butane has the higher boiling point \(\left(-0.5^{\circ} \mathrm{C}\right.\) compared to \(\left.-11.7{ }^{\circ} \mathrm{C}\right)\). Explain.

Benzoic acid, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\), melts at \(122{ }^{\circ} \mathrm{C}\). The density in the liquid state at \(130^{\circ} \mathrm{C}\) is \(1.08 \mathrm{~g} / \mathrm{cm}^{3}\). The density of solid benzoic acid at \(15^{\circ} \mathrm{C}\) is \(1.266 \mathrm{~g} / \mathrm{cm}^{3}\). (a) In which of these two states is the average distance between molecules greater? (b) Explain the difference in densities at the two temperatures in terms of the relative kinetic energies of the molecules.

For each of the following pairs of substances, predict which will have the higher melting point and indicate why: (a) \(\mathrm{Ar}, \mathrm{Xe} ;\) (b) \(\mathrm{SiO}_{2}, \mathrm{CO}_{2} ;\) (c) \(\mathrm{KBr}, \mathrm{Br}_{2} ;\) (d) \(\mathrm{C}_{6} \mathrm{Cl}_{6}, \mathrm{C}_{6} \mathrm{H}_{6}\)

Ethylene glycol \(\left(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\right)\), the major substance in antifreeze, has a normal boiling point of \(198^{\circ} \mathrm{C}\). By comparison, ethyl alcohol \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\right)\) boils at \(78^{\circ} \mathrm{C}\) at atmospheric pressure. Ethylene glycol dimethyl ether \(\left(\mathrm{CH}_{3} \mathrm{OCH}_{2} \mathrm{CH}_{2} \mathrm{OCH}_{3}\right)\) has a normal boiling point of \(83^{\circ} \mathrm{C}\), and ethyl methyl ether \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OCH}_{3}\right)\) has a normal boiling point of \(11^{\circ} \mathrm{C}\). (a) Explain why replacement of a hydrogen on the oxygen by \(\mathrm{CH}_{3}\) generally results in a lower boiling point. (b) What are the major factors responsible for the difference in boiling points of the two ethers?

Describe the intermolecular forces that must be overcome to convert each of the following from a liquid or solid to a gas: (a) \(\mathrm{I}_{2}\), (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\), (c) \(\mathrm{H}_{2}\) Se.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Making Measurements

Read Explanation

Chemistry Branches

Read Explanation

The Earths Atmosphere

Read Explanation

Organic Chemistry

Read Explanation

Kinetics

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.