Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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 \(P_{\mathrm{N}_{2}}=0.751 \mathrm{~atm}\) and \(P_{\mathrm{O}_{2}}=0.208 \mathrm{~atm} .\) 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.

Short Answer

Expert verified
Short answer: (a) To calculate the partial pressure of diethylether, first determine the mass of diethylether using the volume and density (\(m_{diethylether} = V_{diethylether} \times \rho_{diethylether}\)). Convert the mass to moles using the molar mass (\(n_{diethylether} = \frac{m_{diethylether}}{molar\ mass_{diethylether}}\)). Then, use the Ideal Gas Law to calculate the partial pressure (\(P_{diethylether} = \frac{n_{diethylether}RT}{V}\)). (b) To calculate the total pressure in the container, sum up the partial pressures of N2, O2, and diethylether (\(P_{total} = P_{\mathrm{N}_{2}} + P_{\mathrm{O}_{2}} + P_{diethylether}\)).

Step by step solution

01

Calculate the mass of diethylether

First, let's calculate the mass of diethylether using its volume and density: \(m_{diethylether} = V_{diethylether} \times \rho_{diethylether}\) where \(m_{diethylether}\) is the mass of diethylether (in grams), \(V_{diethylether}\) is the volume of the diethylether (5.00 mL), and \(\rho_{diethylether}\) is the density of the diethylether (0.7134 g/mL).
02

Convert the mass of diethylether to moles

To convert the mass of diethylether to moles, we need to use the molar mass of diethylether: Molecular formula of diethylether is \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OC}_{2} \mathrm{H}_{5}\), which gives us a molar mass of \(4 * 12.01 + 10 * 1.01 + 1 * 16 = 74.12 \mathrm{g/mol}\). Now we can calculate the moles of diethylether: \(n_{diethylether} = \frac{m_{diethylether}}{molar\ mass_{diethylether}}\)
03

Calculate the partial pressure of diethylether

Now that we have the moles of diethylether, we can use the Ideal Gas Law to calculate the partial pressure of diethylether: \(P_{diethylether}V = n_{diethylether}RT\) where \(P_{diethylether}\) is the partial pressure of diethylether, V is the volume of the container (6.00 L), \(n_{diethylether}\) is the moles of diethylether calculated in Step 2, R is the ideal gas constant (\(0.08206 \mathrm{L~atm/mol~K}\)), and T is the temperature in Kelvin (35.0°C = 308.15 K). Now we can solve for \(P_{diethylether}\): \(P_{diethylether} = \frac{n_{diethylether}RT}{V}\)
04

Calculate the total pressure in the container

Finally, we can calculate the total pressure in the container by summing up the partial pressures of N2, O2, and diethylether: \(P_{total} = P_{\mathrm{N}_{2}} + P_{\mathrm{O}_{2}} + P_{diethylether}\) With this step by step solution, you should now have all the necessary information to solve for the partial pressure of diethylether (a) and the total pressure in the container (b).

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 Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Pressure
In a mixture of gases, each gas has its own pressure, called a partial pressure. Partial pressure is the pressure a gas would exert if it occupied the entire volume by itself without mixing with other gases. When you add the partial pressures of all gases in a container, you obtain the total pressure inside the container. This is expressed by Dalton's Law of Partial Pressures, which states:
  • \[P_{total} = P_{1} + P_{2} + \ldots + P_{n} \]

The ideal gas law can also help us find the partial pressure of a gas if we know its amount in moles, the volume of the container, and the temperature. For example, in this exercise, diethylether's partial pressure is found using the ideal gas law: \[P_{diethylether} = \frac{n_{diethylether}RT}{V}\] where:
  • \(n_{diethylether}\) is the moles of diethylether
  • \(R\) is the ideal gas constant
  • \(T\) is the temperature in Kelvin
  • \(V\) is the volume of the container
Molar Mass
Molar mass is a fundamental concept in chemistry that tells us the mass of one mole of a substance. It's the sum of the atomic masses of all atoms in the molecule. Molar mass is crucial in converting a substance's mass to its amount in moles.For instance, in this exercise, calculating the molar mass of diethylether involves summing the atomic masses of all atoms present:
  • Diethylether is composed of \(\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OC}_{2}\mathrm{H}_{5}\), which is made up of carbon (C), hydrogen (H), and oxygen (O).
  • Its molar mass is calculated as: \(4 \times 12.01 + 10 \times 1.01 + 1 \times 16 = 74.12 \ \mathrm{g/mol} \)

This molar mass is then used to convert the mass of diethylether to moles, which is a critical step for further calculations.
Density
Density is a measure of how much mass is contained in a given volume. It is commonly expressed in grams per milliliter (g/mL) or grams per cubic centimeter (g/cm³). Density is important because it allows us to convert between mass and volume.In this exercise, we used the density of diethylether to determine its mass from its volume as follows:
  • The volume of diethylether given is \(5.00 \ \mathrm{mL}\).
  • The density is \(0.7134 \ \mathrm{g/mL}\).

The relationship between mass, volume, and density is expressed by the formula:
  • \(m = V \cdot \rho \)

where \(m\) is the mass, \(V\) is the volume, and \(\rho\) is the density. This concept allows us to convert volume to mass, which is a necessary step before converting mass to moles.
Chemical Calculations
Chemical calculations are a core part of problem-solving in chemistry, involving the application of ratios, formulas, and constants to find unknowns like mass, moles, and pressure. This exercise employed several key chemical calculation steps:
  • First, we calculated the mass of diethylether using its volume and density.
  • Then, using the molar mass, we converted this mass to moles.
  • With the moles known, we applied the ideal gas law to find the partial pressure of diethylether.
  • Finally, we summed partial pressures to find the total pressure in the container.

These steps highlight the importance of understanding and applying basic chemical concepts and calculations to solve complex chemistry problems efficiently and accurately.

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

In Sample Exercise 10.16 , we found that one mole of \(\mathrm{Cl}_{2}\) confined to \(22.41 \mathrm{~L}\) at \(0{ }^{\circ} \mathrm{C}\) deviated slightly from ideal behavior. Calculate the pressure exerted by \(1.00 \mathrm{~mol} \mathrm{Cl}_{2}\) confined to a smaller volume, \(5.00 \mathrm{~L}\), at \(25^{\circ} \mathrm{C} .\) (a) First use the ideal-gas equation and (b) then use the van der Waals equation for your calculation. (Values for the van der Waals constants are given in Table \(10.3 .)\) (c) Why is the difference between the result for an ideal gas and that calculated using the van der Waals equation greater when the gas is confined to \(5.00 \mathrm{~L}\) compared to \(22.4 \mathrm{~L} ?\)

(a) How is the law of combining volumes explained by Avogadro's hypothesis? (b) Consider a 1.0 - \(\mathrm{L}\) flask containing neon gas and a 1.5-L flask containing xenon gas. Both gases are at the same pressure and temperature. According to Avogadro's law, what can be said about the ratio of the number of atoms in the two flasks? (c) Will 1 mol of an ideal gas always occupy the same volume at a given temperature and pressure? Explain.

Rank the following gases from least dense to most dense at 1.00 atm and \(298 \mathrm{~K}: \mathrm{SO}_{2}, \mathrm{HBr}, \mathrm{CO}_{2} .\) Explain.

The metabolic oxidation of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},\) in our bodies produces \(\mathrm{CO}_{2},\) which is expelled from our lungs as a gas: $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) $$ (a) Calculate the volume of dry \(\mathrm{CO}_{2}\) produced at body temperature \(\left(37^{\circ} \mathrm{C}\right)\) and 0.970 atm when \(24.5 \mathrm{~g}\) of glucose is consumed in this reaction. (b) Calculate the volume of oxygen you would need, at 1.00 atm and \(298 \mathrm{~K}\), to completely oxidize \(50.0 \mathrm{~g}\) of glucose.

Imagine that the reaction \(2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)\) occurs in a container that has a piston that moves to maintain a constant pressure when the reaction occurs at constant temperature. (a) What happens to the volume of the container as a result of the reaction? Explain. (b) If the piston is not allowed to move, what happens to the pressure as a result of the reaction? [Sections 10.3 and 10.5 ]

See all solutions

Recommended explanations on Chemistry Textbooks

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