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

In the Dumas-bulb technique for determining the molar mass of an unknown liquid, you vaporize the sample of a liquid that boils below \(100^{\circ} \mathrm{C}\) in a boiling-water bath and determine the mass of vapor required to fill the bulb (see drawing, next page). From the following data, calculate the molar mass of the unknown liquid: mass of unknown vapor, \(1.012 \mathrm{~g}\); volume of bulb, \(354 \mathrm{~cm}^{3} ;\) pressure, 742 torr; temperature, \(99^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The molar mass of the unknown liquid is approximately \(28.4~\text{g/mol}\), which is calculated using the ideal gas law, converting units, and finding the number of moles before dividing the mass of the vapor by the moles.

Step by step solution

01

Convert the given data to appropriate units

Before we use the ideal gas law formula, we need to make sure the given values are in the correct units. The pressure should be in atmospheres, the volume should be in liters, and the temperature should be in Kelvin. Given values: - Mass of unknown vapor, \(m = 1.012~\text{g}\) - Volume of bulb, \(V = 354~\text{cm}^3\) - Pressure, \(P = 742~\text{torr}\) - Temperature, \(T = 99^\circ \text{C}\) Converting the units, we get: - Volume in liters: \(V = 354~\text{cm}^3 \times \frac{1~\text{L}}{1000~\text{cm}^3} = 0.354~\text{L}\) - Pressure in atmospheres: \(P = 742~\text{torr} \times \frac{1~\text{atm}}{760~\text{torr}} = 0.9763~\text{atm}\) - Temperature in Kelvin: \(T = 99^\circ \text{C} + 273.15 = 372.15~\text{K}\)
02

Use the ideal gas law to find the number of moles

Now that we have the necessary values in the appropriate units, we can use the ideal gas law formula, which is: \(PV = nRT\) Where: - P is the pressure in atmospheres. - V is the volume in liters. - n is the number of moles of gas. - R is the ideal gas constant (\(0.0821~\text{L·atm/mol·K}\)) - T is the temperature in Kelvin. We need to find the value of n, so we can rewrite the formula as: \(n = \frac{PV}{RT}\) Substituting the given values, we get: \(n = \frac{(0.9763~\text{atm})(0.354~\text{L})}{(0.0821~\text{L·atm/mol·K})(372.15~\text{K})} \approx 0.0356~\text{mol}\)
03

Calculate the molar mass of the unknown liquid

We now have the number of moles of the vapor (n) and the mass of the vapor (m). We can use this information to calculate the molar mass (M) of the unknown liquid using the formula: \(M = \frac{\text{mass}}{\text{moles}}\) Using the given mass and the calculated moles, we get: \(M = \frac{1.012~\text{g}}{0.0356~\text{mol}} \approx 28.4~\text{g/mol}\) The molar mass of the unknown liquid is approximately \(28.4~\text{g/mol}\).

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.

Dumas-bulb technique
The Dumas-bulb technique is a classic laboratory method for determining the molar mass of a volatile liquid. The process involves vaporizing a known mass of a liquid in a flask called a Dumas bulb, which is then heated to boil off the liquid. As the liquid vaporizes, it displaces air inside the bulb until only the vapor of the liquid remains. Once the vapor has filled the bulb, it exerts a pressure equal to the atmospheric pressure, if the bulb is open to the atmosphere, or a known pressure otherwise.

A key point in this process is that all air must be expelled from the bulb to ensure that the pressure measured is solely due to the vapor of the liquid. Assume ideal behavior of the gas, the parameters such as the volume of the bulb, the pressure of the vapor, and the temperature of the surrounding environment are used to determine the amount of gas present. Since the mass of the vapor is known, this allows for direct calculation of the molar mass. It's essential to take the measurements when the vapor is in equilibrium with its liquid to prevent the loss of material and obtain accurate results.
Ideal gas law
The ideal gas law is a fundamental equation in chemistry that relates the pressure, volume, temperature, and number of moles of an ideal gas. Expressed as PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the temperature in Kelvin. Ideal gases are hypothetical gases whose molecules occupy negligible space and have no intermolecular forces.

Because real gases exhibit non-ideal behavior, the ideal gas law is most accurate under conditions of low pressure and high temperature. In the exercise solution, the ideal gas law helps calculate the number of moles of vapor present. With the assumption that the vapor behaves as an ideal gas, the mass and the volume of the gas allows for a straightforward determination of its molar mass. To apply the ideal gas law effectively, remember that all units must be consistent, which is why unit conversion is a critical step before solving.
Unit conversion
Unit conversion is crucial in scientific calculations to ensure that equations are dimensionally consistent and that quantities are accurately compared. In chemistry, particularly in gas laws, it is common to work with standard units of liters for volume, Kelvin for temperature, and atmospheres for pressure. If measurements are taken in other units, they must be converted before using them in formulas.

For example, as done in the exercise, volume is converted from cubic centimeters to liters (1 L = 1000 cm3), pressure is converted from torr to atmospheres (1 atm = 760 torr), and temperature is converted from Celsius to Kelvin (K = °C + 273.15). Remember that incorrect or inconsistent units can lead to substantial errors in the final result. Always be meticulous with unit conversion to ensure the integrity of your calculations.

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

Which of the following statements best explains why a closed balloon filled with helium gas rises in air? (a) Helium is a monatomic gas, whereas nearly all the molecules that make up air, such as nitrogen and oxygen, are diatomic. (b) The average speed of helium atoms is higher than the average speed of air molecules, and the higher speed of collisions with the balloon walls propels the balloon upward. (c) Because the helium atoms are of lower mass than the average air molecule, the helium gas is less dense than air. The mass of the balloon is thus less than the mass of the air displaced by its volume. (d) Because helium has a lower molar mass than the average air molecule, the helium atoms are in faster motion. This means that the temperature of the helium is higher than the air temperature. Hot gases tend to rise.

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.

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 ]

(a) Calculate the number of molecules in a deep breath of air whose volume is \(2.25 \mathrm{~L}\) at body temperature, \(37{ }^{\circ} \mathrm{C},\) and a pressure of 735 torr. (b) The adult blue whale has a lung capacity of \(5.0 \times 10^{3} \mathrm{~L} .\) Calculate the mass of air (assume an average molar mass \(28.98 \mathrm{~g} / \mathrm{mol}\) ) contained in an adult blue whale's lungs at \(0.0{ }^{\circ} \mathrm{C}\) and 1.00 atm, assuming the air behaves ideally.

A piece of dry ice (solid carbon dioxide) with a mass of \(5.50 \mathrm{~g}\) is placed in a 10.0 - \(\mathrm{L}\) vessel that already contains air at 705 torr and \(24^{\circ} \mathrm{C}\). After the carbon dioxide has totally vaporized, what is the partial pressure of carbon dioxide and the total pressure in the container at \(24{ }^{\circ} \mathrm{C} ?\)

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