Ch.12 - SolutionsWorksheetSee all chapters
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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

Solution: Find the mass of urea (CH4N2O) needed to prepare 50.1 g of a solution in water in which the mole fraction of urea is 7.66×10−2.

Solution: Find the mass of urea (CH4N2O) needed to prepare 50.1 g of a solution in water in which the mole fraction of urea is 7.66×10−2.

Problem

Find the mass of urea (CH4N2O) needed to prepare 50.1 g of a solution in water in which the mole fraction of urea is 7.66×10−2.

Solution

Mole fraction is the ratio of the number of moles of one substance over the total number of moles that are presentThis can be presented as below where the mole ratio of urea in the solution can be found.


where,

Xurea is the mole ratio of urea present in the solution and nurea and nwater are the number of moles of urea and water that is present in the solution.


As we know, the mole fraction of urea in the solution to be made is to equal to, 

χurea = 7.66 x 10-2  0.0766


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