Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 11: Problem 49

The solubility of nitrogen in water is \(8.21 \times 10^{-4} \mathrm{~mol} / \mathrm{L}\) at \(0^{\circ} \mathrm{C}\) when the \(\mathrm{N}_{2}\) pressure above water is \(0.790 \mathrm{~atm} .\) Calculate the Henry's law constant for \(\mathrm{N}_{2}\) in units of \(\mathrm{mol} / \mathrm{L} \cdot \mathrm{atm}\) for Henry's law in the form \(C=k P\), where \(C\) is the gas concentration in mol/L. Calculate the solubility of \(\mathrm{N}_{2}\) in water when the partial pressure of nitrogen above water is \(1.10 \mathrm{~atm}\) at \(0^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The Henry's law constant for nitrogen in units of mol/L·atm is approximately \(1.04 \times 10^{-3} \mathrm{~mol}/\mathrm{L} \cdot \mathrm{atm}\). The solubility of nitrogen in water when the partial pressure of nitrogen above water is \(1.10 \mathrm{~atm}\) at \(0^{\circ} \mathrm{C}\) is approximately \(1.14 \times 10^{-3} \mathrm{~mol}/\mathrm{L}\).

Step by step solution

01

Write down the given information

We are given: - The solubility of nitrogen in water at \(0^{\circ}\mathrm{C}\) and \(0.790 \mathrm{~atm}\) is \(8.21 \times 10^{-4} \mathrm{~mol}/\mathrm{L}\). - The partial pressure of nitrogen above water is \(1.10 \mathrm{~atm}\) at \(0^{\circ} \mathrm{C}\). We need to determine the Henry's law constant and solubility of nitrogen in water at the new partial pressure.
02

Use Henry's law formula

According to Henry's law, the solubility of a gas (C) is directly proportional to the partial pressure of the gas (P) when the temperature remains constant. The formula for Henry's law is: \[C = kP\] where \(C\) is the gas concentration in mol/L, \(k\) is the Henry's law constant in units of mol/L·atm, and \(P\) is the partial pressure of the gas in atm.
03

Calculate the Henry's law constant (k)

Using the given information, we can calculate the Henry's law constant (k) as follows: \[k = \dfrac{C}{P} = \dfrac{8.21 \times 10^{-4} \mathrm{~mol}/\mathrm{L}}{0.790 \mathrm{~atm}}\] Now, compute the value of k: \[k \approx 1.04 \times 10^{-3} \mathrm{~mol}/\mathrm{L} \cdot \mathrm{atm}\]
04

Calculate the solubility of nitrogen in water at the new partial pressure

According to Henry's law, we have the relationship: \[C = kP\] Now, plug in the values of \(k\) and the new partial pressure (\(P = 1.10 \mathrm{~atm}\)): \[C = (1.04 \times 10^{-3} \mathrm{~mol}/\mathrm{L} \cdot \mathrm{atm})(1.10 \mathrm{~atm})\] Compute the solubility of nitrogen (C) in water at the new partial pressure: \[C \approx 1.14 \times 10^{-3} \mathrm{~mol}/\mathrm{L}\] So, the solubility of nitrogen in water when the partial pressure of nitrogen above water is \(1.10 \mathrm{~atm}\) at \(0^{\circ} \mathrm{C}\) is approximately \(1.14 \times 10^{-3} \mathrm{~mol}/\mathrm{L}\).

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

The weak electrolyte \(\mathrm{NH}_{3}(g)\) does not obey Henry's law. Why? \(\mathrm{O}_{2}(g)\) obeys Henry's law in water but not in blood (an aqueous solution). Why?

A solution is prepared by mixing \(25 \mathrm{~mL}\) pentane \(\left(\mathrm{C}_{5} \mathrm{H}_{12}, d=\right.\) \(0.63 \mathrm{~g} / \mathrm{cm}^{3}\) ) with \(45 \mathrm{~mL}\) hexane \(\left(\mathrm{C}_{6} \mathrm{H}_{14}, d=0.66 \mathrm{~g} / \mathrm{cm}^{3}\right)\). Assuming that the volumes add on mixing, calculate the mass percent, mole fraction, molality, and molarity of the pentane.

You drop an ice cube (made from pure water) into a saltwater solution at \(0^{\circ} \mathrm{C}\). Explain what happens and why.

A \(1.60-\mathrm{g}\) sample of a mixture of naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\) and anthracene \(\left(\mathrm{C}_{14} \mathrm{H}_{10}\right)\) is dissolved in \(20.0 \mathrm{~g}\) benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{\mathrm{b}}\right)\). The freezing point of the solution is \(2.81^{\circ} \mathrm{C}\). What is the composition as mass percent of the sample mixture? The freezing point of benzene is \(5.51^{\circ} \mathrm{C}\) and \(K_{\mathrm{f}}\) is \(5.12^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}\).

A solution of sodium chloride in water has a vapor pressure of \(19.6\) torr at \(25^{\circ} \mathrm{C}\). What is the mole fraction of solute particles in this solution? What would be the vapor pressure of this solution at \(45^{\circ} \mathrm{C} ?\) The vapor pressure of pure water is \(23.8\) torr at \(25^{\circ} \mathrm{C}\) and \(71.9\) torr at \(45^{\circ} \mathrm{C}\), and assume sodium chloride exists as \(\mathrm{Na}^{+}\) and \(\mathrm{Cl}^{-}\) ions in solution.
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Kinetics

Read Explanation

Chemical Analysis

Read Explanation

Physical Chemistry

Read Explanation

Organic Chemistry

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