Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 111

A glass vessel fitted with a stopcock valve has a mass of $337.428 \mathrm{~g}\( when evacuated. When filled with \)\mathrm{Ar}$, it has a mass of \(339.854 \mathrm{~g}\). When evacuated and refilled with a mixture of Ne and Ar, under the same conditions of temperature and pressure, it has a mass of \(339.076 \mathrm{~g} .\) What is the mole percent of Ne in the gas mixture?

Short Answer

Expert verified
The mole percent of Ne in the gas mixture is approximately 38.80%.

Step by step solution

01

Find the mass of Ar when only filled with Ar

Subtract the mass of the evacuated glass vessel from the mass of the vessel when filled with Ar to determine the mass of Ar: Mass of Ar_only = Mass of filled vessel with Ar - Mass of evacuated vessel Mass of Ar_only = \(339.854 \mathrm{~g}\) - \(337.428 \mathrm{~g}\) Mass of Ar_only = 2.426 g
02

Calculate moles of Ar when only filled with Ar

Now, we need to find the moles of Ar when filled only with Ar. To do this, use the molar mass of Ar (39.95 g/mol): Moles of Ar_only = Mass of Ar_only / Molar mass of Ar Moles of Ar_only = \(2.426 \mathrm{~g}\) / \(39.95 \mathrm{~g/mol}\) Moles of Ar_only = 0.06073 mol
03

Find the mass of the gas mixture

Subtract the mass of the evacuated glass vessel from the mass of the vessel when filled with the mixture of Ne and Ar: Mass of gas_mixture = Mass of filled vessel with gas mixture - Mass of evacuated vessel Mass of gas_mixture = \(339.076 \mathrm{~g}\) - \(337.428 \mathrm{~g}\) Mass of gas_mixture = 1.648 g
04

Assume that the mole amount of Ar is the same in both cases

Since the conditions of temperature and pressure are the same, the moles of Ar in the gas mixture should be the same as when filled with only Ar: Moles of Ar_mixture = 0.06073 mol
05

Calculate the mass of Ar in the gas mixture

Now, we need to find the mass of Ar in the gas mixture: Mass of Ar_mixture = Moles of Ar_mixture * Molar mass of Ar Mass of Ar_mixture = \(0.06073 \mathrm{~mol}\) * \(39.95 \mathrm{~g/mol}\) Mass of Ar_mixture = 2.425 g
06

Calculate the mass and moles of Ne in the gas mixture

Subtract the mass of Ar_mixture from the total mass of the gas mixture to find the mass of Ne: Mass of Ne = Mass of gas_mixture - Mass of Ar_mixture Mass of Ne = \(1.648 \mathrm{~g}\) - \(2.425 \mathrm{~g}\) Mass of Ne = -0.777 g Calculate the moles of Ne using the molar mass of Ne (20.18 g/mol): Moles of Ne = Mass of Ne / Molar mass of Ne Moles of Ne = \(-0.777 \mathrm{~g}\) / \(20.18 \mathrm{~g/mol}\) Moles of Ne = 0.03847 mol
07

Calculate the mole percent of Ne in the gas mixture

To find the mole percent of Ne in the gas mixture, use the following formula: Mole percent of Ne = (Moles of Ne / (Moles of Ar_mixture + Moles of Ne)) * 100 Mole percent of Ne = \((0.03847 \mathrm{~mol}\) / (\(0.06073 \mathrm{~mol}\) + \(0.03847 \mathrm{~mol}\))) * 100 Mole percent of Ne = \(38.80\%\) The mole percent of Ne in the gas mixture is approximately 38.80%.

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

Which one or more of the following statements are true? (a) \(\mathrm{O}_{2}\) will effuse faster than \(\mathrm{Cl}_{2}\). (b) Effusion and diffusion are different names for the same process. (c) Perfume molecules travel to your nose by the process of effusion. (d) The higher the density of a gas, the shorter the mean free path.

Chlorine dioxide gas \(\left(\mathrm{ClO}_{2}\right)\) is used as a commercial bleaching agent. It bleaches materials by oxidizing them. In the course of these reactions, the \(\mathrm{ClO}_{2}\) is itself reduced. (a) What is the Lewis structure for \(\mathrm{ClO}_{2} ?\) (b) Why do you think that \(\mathrm{ClO}_{2}\) is reduced so readily? (c) When a \(\mathrm{ClO}_{2}\), molecule gains an electron, the chlorite ion, \(\mathrm{ClO}_{2}^{-},\) forms. Draw the Lewis structure for \(\mathrm{ClO}_{2}^{-}\). (d) Predict the \(\mathrm{O}-\mathrm{Cl}-\mathrm{O}\) bond angle in the \(\mathrm{ClO}_{2}^{-}\) ion. (e) One method of preparing \(\mathrm{ClO}_{2}\) is by the reaction of chlorine and sodium chlorite: $$\mathrm{Cl}_{2}(g)+2 \mathrm{NaClO}_{2}(s) \longrightarrow 2 \mathrm{ClO}_{2}(g)+2 \mathrm{NaCl}(s)$$ If you allow \(15.0 \mathrm{~g}\) of \(\mathrm{NaClO}_{2}\) to react with $2.00 \mathrm{~L}\( of chlorine gas at a pressure of \)152.0 \mathrm{kPa}$ at \(21^{\circ} \mathrm{C},\) how many grams of \(\mathrm{ClO}_{2}\) can be prepared?

A scuba diver's tank contains \(2.50 \mathrm{~kg}\) of \(\mathrm{O}_{2}\) compressed into a volume of \(11.0 \mathrm{~L}\). (a) Calculate the gas pressure inside the tank at \(10^{\circ} \mathrm{C}\). (b) What volume would this oxygen occupy at \(25^{\circ} \mathrm{C}\) and \(101.33 \mathrm{kPa} ?\)

(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?

Assume that a single cylinder of an automobile engine has a volume of $524 \mathrm{~cm}^{3}\(. (a) If the cylinder is full of air at \)74^{\circ} \mathrm{C}\( and \)99.3 \mathrm{kPa}\(, how many moles of \)\mathrm{O}_{2}$ are present? (The mole fraction of \(\mathrm{O}_{2}\) in dry air is \(0.2095 .\) ) (b) How many grams of \(\mathrm{C}_{8} \mathrm{H}_{18}\) could be combusted by this quantity of \(\mathrm{O}_{2}\), assuming complete combustion with formation of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O} ?\)
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Chemical Reactions

Read Explanation

Nuclear Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Inorganic Chemistry

Read Explanation

Physical 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