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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: Assuming 100% dissociation, calculate the freezing point and boiling point of 2.96 m Na2SO4(aq).

Solution: Assuming 100% dissociation, calculate the freezing point and boiling point of 2.96 m Na2SO4(aq).

Problem

Assuming 100% dissociation, calculate the freezing point and boiling point of 2.96 m Na2SO4(aq).


Solution

We’re being asked to determine the freezing point and boiling point of an aqueous Na2SO4 solution. Aqueous means that Na2SO4 was dissolved in water.


When calculating the freezing point of a solution, we’re going to use the Freezing Point Depression equation. On the other hand, when calculating for the boiling point of a solution we’re going to use the Boiling Point Elevation equation.


Freezing Point Depression:

∆Tf = change in freezing point = Tf pure solvent –Tf solution
Kf = freezing point depression constant
i = van' t Hoff factor of the solute = no. of ions
m = molality

Boiling Point Elevation:

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