Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 53

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. From the following data, calculate the molar mass of the unknown liquid: mass of unknown vapor, 1.012 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 25.18 g/mol, calculated using the given data and the Ideal Gas Law equation.

Step by step solution

01

Convert units of given data

Before we proceed to use the Ideal Gas Law equation, we should make sure that all the data is in the correct units. The ideal gas constant (R) is given in units of L atm/mol K. So, we need to convert the volume to liters, temperature to Kelvin, and pressure to atm. 1. Volume conversion: Volume = 354 cm³ = 0.354 L (since there are 1,000 cm³ in a liter) 2. Temperature conversion: Temperature = 99°C = 99 + 273.15 = 372.15 K 3. Pressure conversion: Pressure = 742 torr = (742/760) atm ≈ 0.976 atm (since there are 760 torr in 1 atm) Now we have the required units for the Ideal Gas Law equation: Volume (V) = 0.354 L Temperature (T) = 372.15 K Pressure (P) = 0.976 atm
02

Calculate the number of moles using the Ideal Gas Law equation

Now that we have all the data in the appropriate units, we can use the Ideal Gas Law equation to calculate the number of moles (n). Rearranging the Ideal Gas Law equation for n: \(n = \frac{PV}{RT}\) Using the given data: n = (0.976 atm) * (0.354 L) / (0.0821 L atm/mol K) * (372.15 K) n ≈ 0.0402 moles
03

Calculate the molar mass of the unknown liquid

Now that we have the number of moles of the unknown vapor, we can calculate the molar mass of the unknown liquid using the formula: Molar mass = mass of unknown vapor/number of moles Molar mass = (1.012 g) / (0.0402 moles) Molar mass ≈ 25.18 g/mol The molar mass of the unknown liquid is approximately 25.18 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.

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

\((\mathbf{a})\) Amonton's law expresses the relationship between pressure and temperature. Use Charles's law and Boyle's law to derive the proportionality relationship between \(P\) and \(T\) . \((\mathbf{b})\) If a car tire is filled to a pressure of 32.0 \(\mathrm{lb} / \mathrm{in.}^{2}\) (psi) measured at \(75^{\circ} \mathrm{F},\) what will be the tire pressure if the tires heat up to \(120^{\circ} \mathrm{F}\) during driving?

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

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. Which of the following statements describes how the volume of the container changes due to the reaction: (a) the volume increases by \(50 \%,(\mathbf{b})\) the volume increases by \(33 \%,\) (c) the volume remains constant, (d) the volume decreases by \(33 \%,(\) e) the volume decreases by 50\(\% .[\) Sections 10.3 and 10.4\(]\)

Hydrogen has two naturally occurring isotopes, \(^{1} \mathrm{H}\) and \(^{2} \mathrm{H}\) . Chlorine also has two naturally occurring isotopes, 35 \(\mathrm{Cl}\) and37 Cl. Thus, hydrogen chloride gas consists of four distinct types of molecules: \(^{1} \mathrm{H}^{35} \mathrm{Cl},^{2} \mathrm{H}^{37} \mathrm{Cl},^{2} \mathrm{H}^{35} \mathrm{Cl},\) and \(^{2} \mathrm{H}^{37} \mathrm{Cl}\) Place these four molecules in order of increasing rate of effusion.

An herbicide is found to contain only \(\mathrm{C}, \mathrm{H}, \mathrm{N},\) and Cl. The complete combustion of a 100.0 -mg sample of the herbicide in excess oxygen produces 83.16 \(\mathrm{mL}\) of \(\mathrm{CO}_{2}\) and 73.30 \(\mathrm{mL}\) of \(\mathrm{H}_{2} \mathrm{O}\) vapor expressed at STP. A separate analysis shows that the sample also contains 16.44 \(\mathrm{mg}\) of Cl. (a) Determine the percentage of the composition of the substance. ( b) Calculate its empirical formula. (c) What other information would you need to know about this compound to calculate its true molecular formula?
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Making Measurements

Read Explanation

Chemical Reactions

Read Explanation

Chemical Analysis

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