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 \(\mathrm{mol} / \mathrm{L} \cdot\) 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 N2 is approximately \(1.0392 \times 10^{-3}\,\mathrm{mol/L\cdot atm}\), and the solubility of N2 in water when the partial pressure is 1.10 atm at 0°C is approximately \(1.1431 \times 10^{-3}\,\mathrm{mol/L}\).

Step by step solution

01

Understand Henry's Law and the given variables

Henry's Law can be written as \(C = kP\), where \(C\) is the gas concentration in mol/L, P is the pressure in atm, and k is the Henry's Law constant. We are given the solubility of N2 in water at 0°C when the N2 pressure above water is 0.790 atm, which is \(8.21 \times 10^{-4}\) mol/L.
02

Calculate Henry's Law constant for N2

We can find the Henry's Law constant by rearranging the equation: \[k = \frac{C}{P}\] Use the given values for solubility and pressure to calculate k: \[k = \frac{8.21 \times 10^{-4}\,\mathrm{mol/L}}{0.790\,\mathrm{atm}}\]
03

Simplify the expression

Now, divide \(8.21 \times 10^{-4}\) by 0.790: \[k \approx 1.0392 \times 10^{-3}\, \mathrm{mol/L\cdot atm}\] So, the Henry's Law constant for N2 is approximately \(1.0392 \times 10^{-3}\,\mathrm{mol/L\cdot atm}\).
04

Use Henry's Law constant to calculate the solubility of N2 for the given pressure

We are asked to find the solubility of N2 when the partial pressure above water is 1.10 atm at 0°C. Using Henry's Law: \[C = kP\] Substitute the values of k and P: \[C \approx (1.0392 \times 10^{-3}\,\mathrm{mol/L\cdot atm})(1.10\,\mathrm{atm})\]
05

Compute the solubility of N2 for 1.10 atm pressure

Now, multiply \(1.0392 \times 10^{-3}\) by 1.10: \[C \approx 1.1431 \times 10^{-3}\,\mathrm{mol/L}\] So, the solubility of N2 in water when the partial pressure is 1.10 atm at 0°C is approximately \(1.1431 \times 10^{-3}\,\mathrm{mol/L}\).

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

Explain the terms isotonic solution, crenation, and hemolysis.

The high melting points of ionic solids indicate that a lot of energy must be supplied to separate the ions from one another. How is it possible that the ions can separate from one another when soluble ionic compounds are dissolved in water, often with essentially no temperature change?

Using the phase diagram for water and Raoult's law, explain why salt is spread on the roads in winter (even when it is below freezing).

At a certain temperature, the vapor pressure of pure benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) is \(0.930\) atm. A solution was prepared by dissolving \(10.0\) \(\mathrm{g}\) of a nondissociating, nonvolatile solute in \(78.11 \mathrm{~g}\) of benzene at that temperature. The vapor pressure of the solution was found to be \(0.900\) atm. Assuming the solution behaves ideally, determine the molar mass of the solute.

Pentane \(\left(\mathrm{C}_{5} \mathrm{H}_{12}\right)\) and hexane \(\left(\mathrm{C}_{6} \mathrm{H}_{14}\right)\) form an ideal solution. At \(25^{\circ} \mathrm{C}\) the vapor pressures of pentane and hexane are 511 and 150\. torr, respectively. A solution is prepared by mixing 25 mL pentane (density, \(0.63 \mathrm{~g} / \mathrm{mL}\) ) with \(45 \mathrm{~mL}\) hexane (density, \(0.66 \mathrm{~g} / \mathrm{mL})\) a. What is the vapor pressure of the resulting solution? b. What is the composition by mole fraction of pentane in the vapor that is in equilibrium with this solution?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Analysis

Read Explanation

Organic Chemistry

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