Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 11: Q11.3-24E (page 647)

The Henry’s law constant for \({\bf{O}}_2\)is \({\bf{1}}{\bf{.3 \times 1}}{{\bf{0}}^{{\bf{ - 3}}}}{\bf{ M at}}{{\bf{m}}^{{\bf{ - 1}}}}\)at 25 °C. What mass of oxygen would be dissolved in a 40 L aquarium at 25 °C, assuming an atmospheric pressure of 1.00 atm, and that the partial pressure of \({\bf{O}}_2\)is 0.21 atm?

Short Answer

Expert verified

The mass of oxygen dissolved in a 40-L aquarium at 25 °C is 0.35g. \(\)

Step by step solution

01

Definition of Henry’s Law

Henry’s law is a gas law that states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid.

02

Explanation

This problem requires the application of Henry’s law.

The governing equation is:

Cg = kPg

Cg= Concentration of a dissolved gas

k = Henry's Law Constant

Pg= Partial pressure of the gas

Cg = kPg

Cg = 0.0013Matm-1x 0.21atm =2.7 x 10-4molL-1

The total amount =2.7 x 10-4molL x 40L =0.0108mol

The mass of oxygen =0.0108mol x 32gmol-1=0.346g

Mass of Oxygen with two Significant Figures = 0.35g

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

Calculate the molality of each of the following solutions:

  1. 583 g of \({{\bf{H}}_{\bf{2}}}{\bf{S}}{{\bf{O}}_{\bf{4}}}\) in 1.50 kg of water—the acid solution used in an automobile battery
  2. 0.86 g of NaCl in 1.00 × 102 g of water—a solution of sodium chloride for intravenous injection
  3. 46.85 g of codeine, \({{\bf{C}}_{{\bf{18}}}}{{\bf{H}}_{{\bf{21}}}}{\bf{N}}{{\bf{O}}_{\bf{3}}}\), in 125.5 g of ethanol, \({{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{5}}}{\bf{OH}}\)
  4. 25 g of I2 in 125 g of ethanol, \({{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{5}}}{\bf{OH}}\).

What is the freezing point of a solution of 9.04 g of\({\bf{I_2}}\)in 75.5 g of benzene?

  1. Outline the steps necessary to answer the following question.
  2. Answer the question.

Why does 1 mole of sodium chloride depress the freezing point of 1 kg of water almost twice as much as 1 mole of glycerine?

Convert a volume of 9.345 qt to liters.

Explain why the ions \(N{a^ + }\) and \(C{l^ - }\) are strongly solvated in water but not in hexane, a solvent composed of nonpolar molecules.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Kinetics

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Chemical Analysis

Read Explanation

The Earths Atmosphere

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