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, \(1.012 \mathrm{~g}\); volume of bulb, \(354 \mathrm{~cm}^{3}\); pressure, \(98.93 \mathrm{kPa} ;\) temperature, $99{ }^{\circ} \mathrm{C}$.

Short Answer

Expert verified
The molar mass of the unknown liquid can be calculated using the Ideal Gas Law and the given data: mass of unknown vapor \(1.012 \: g\), volume of bulb \(354 \: cm^3\), pressure \(98.93 \: kPa\), and temperature \(99^\circ C\). First, convert the temperature to Kelvin (\(372 K\)) and the volume to liters (\(0.354 L\)). Then, calculate the number of moles (n) using the Ideal Gas Law: \(n \approx 0.0451 \: moles\). Finally, calculate the molar mass (M) using the formula \(M = \frac{mass}{moles}\), which gives \(M \approx 22.43 \: g/mol\).

Step by step solution

01

Convert the given data to appropriate units

We need to convert the temperature given in Celsius to Kelvin and the volume given in cm³ to L: Temperature in Kelvin: \(T(K) = T(^\circ C) + 273\) Volume in Liters: \(V(L) = \frac{V(cm^3)}{1000}\)
02

Calculate the Temperature in Kelvin

Given temperature, \(T = 99^\circ C\) \(T(K) = 99 + 273 = 372 K\)
03

Calculate the Volume in Liters

Given volume, \(V= 354 cm^3\) \(V(L) = \frac{354}{1000} = 0.354 L\)
04

Calculate the number of moles using the Ideal Gas Law

Now, we will use the Ideal Gas Law to find the number of moles (n): \(PV = nRT\) Rearrange the equation to solve for n: \(n = \frac{PV}{RT}\) Plug in the given values: \(n = \frac{(98.93 kPa)(0.354 L)}{(8.314 J/(mol·K))(372 K)}\)
05

Convert pressure from kPa to Pa

1 kPa = \(10^3\) Pa Pressure, \(P = 98.93 kPa \times 10^3 = 98,930 Pa\)
06

Calculate the number of moles (n)

Plug in the converted values: \(n = \frac{(98,930 Pa)(0.354 L)}{(8.314 J/(mol·K))(372 K)}\) \(n \approx 0.0451\: moles\)
07

Calculate Molar Mass

Now, we will use the formula for molar mass (M): \(M = \frac{mass}{moles}\) Given mass of the unknown vapor, \(m = 1.012\: g\) Molar Mass, \(M = \frac{1.012\: g}{0.0451\: moles} \approx 22.43\: g/mol\) The molar mass of the unknown liquid is approximately 22.43 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

(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 \(220.6 \mathrm{kPa}\) measured at \(24^{\circ} \mathrm{C}\), what will be the tire pressure if the tires heat up to \(49^{\circ} \mathrm{C}\) during driving?

The temperature of a \(5.00-\mathrm{L}\) container of \(\mathrm{N}_{2}\) gas is increased from \(20^{\circ} \mathrm{C}\) to \(250^{\circ} \mathrm{C}\). If the volume is held constant, predict qualitatively how this change affects the following: (a) the average kinetic energy of the molecules; (b) the rootmean- square speed of the molecules; (c) the strength of the impact of an average molecule with the container walls; \(\mathbf{d}\) ) the total number of collisions of molecules with walls per second.

A neon sign is made of glass tubing whose inside diameter is $3.0 \mathrm{~cm}\( and length is \)10.0 \mathrm{~m}$. If the sign contains neon at a pressure of \(265 \mathrm{~Pa}\) at \(30^{\circ} \mathrm{C}\), how many grams of neon are in the sign? (The volume of a cylinder is \(\pi r^{2} h\).)

Calcium hydride, \(\mathrm{CaH}_{2}\), reacts with water to form hydrogen gas: $$\mathrm{CaH}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+2 \mathrm{H}_{2}(g)$$ This reaction is sometimes used to inflate life rafts, weather balloons, and the like, when a simple, compact means of generating \(\mathrm{H}_{2}\) is desired. How many grams of \(\mathrm{CaH}_{2}\) are needed to generate $145 \mathrm{~L}\( of \)\mathrm{H}_{2}\( gas if the pressure of \)\mathrm{H}_{2}$ is \(110 \mathrm{kPa}\) at \(21^{\circ} \mathrm{C} ?\)

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 greater than the average speed of air molecules, and the greater 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 greater than the air temperature. Hot gases tend to rise.
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Organic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Physical 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