Ch.12 - SolutionsWorksheetSee all chapters
All Chapters
Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch. 17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds
Problem

In the flask diagrammed below, the left bulb contains 3.0L of neon at a pressure of 2.0 atm. The right bulb contains 2.0L of argon at a pressure of 1.0 atm. What is the mole fraction of neon and argon in the final mixture after the valve is opened and a uniform pressure is reached at constant temperature (25°C).


(a) Ne 0.33, Ar 0.67

(b) Ne 0.25, Ar 0.75

(c) Ne 0.67, Ar 0.33

(d) Ne 0.92, Ar 0.82

(e) Ne 0.75, Ar 0.25

Solution

We are asked to calculate the mole fractions of each of the gases. 


Mole Fraction (X) relates the moles of components and the total moles:

mole fraction (X)=mole componentmole total


Now to get the total moles we have:

ntotal = nNe + nAr


So for the mole fraction of Ne and Ar we have: 

XNe = nNenNe+nArXAr = nArnNe+nAr


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