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

Solution: Calculate the mass of ethylene glycol (C2H6O2, molar mass = 62.07 g/mol) that must be added to 1.00 kg of ethanol (C2H5OH, molar mass = 46.07 g/mol) to reduce its vapor pressure by 10.0 torr at 35 °C.

Solution: Calculate the mass of ethylene glycol (C2H6O2, molar mass = 62.07 g/mol) that must be added to 1.00 kg of ethanol (C2H5OH, molar mass = 46.07 g/mol) to reduce its vapor pressure by 10.0 torr at 35 °C.

Problem

Calculate the mass of ethylene glycol (C2H6O2, molar mass = 62.07 g/mol) that must be added to 1.00 kg of ethanol (C2H5OH, molar mass = 46.07 g/mol) to reduce its vapor pressure by 10.0 torr at 35 °C. The vapor pressure of pure ethanol at 35 °C is 100 torr.

Solution

Given:

• molar mass ethylene glycol = 62.07 g/mol
• molar mass ethanol = 46.07 g/mol
• mass ethanol = 1.00 kg
• vapor pressure pure ethanol = 100 torr
• vapor pressure of solution → vapor pressure solvent reduced by 10 torr

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