Domanda del libro di testo
Use the van't Hoff factors in Table 13.9 to calculate each colligative property: a. the melting point of a 0.100 m iron(III) chloride solution
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Ch.14 - Solutions
Tro6th EditionChemistry: A Molecular ApproachISBN: 9780137832217
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TOC.tableOfContents
- Ch.1 - Matter, Measurement & Problem Solving118
- Ch.2 - Atoms & Elements106
- Ch.3 - Molecules and Compounds131
- Ch.4 - Chemical Reactions and Chemical Quantities56
- Ch.5 - Introduction to Solutions and Aqueous Solutions82
- Ch.6 - Gases114
- Ch.7 - Thermochemistry96
- Ch.8 - The Quantum-Mechanical Model of the Atom60
- Ch.9 - Periodic Properties of the Elements90
- Ch.10 - Chemical Bonding I: The Lewis Model89
- Ch.11 - Chemical Bonding II: Molecular Shapes, VSEPR & MO Theory85
- Ch.12 - Liquids, Solids & Intermolecular Forces51
- Ch.13 - Solids & Modern Materials47
- Ch.14 - Solutions90
- Ch.15 - Chemical Kinetics115
- Ch.16 - Chemical Equilibrium68
- Ch.17 - Acids and Bases136
- Ch.18 - Aqueous Ionic Equilibrium153
- Ch.19 - Free Energy & Thermodynamics93
- Ch.20 - Electrochemistry106
- Ch.21 - Radioactivity & Nuclear Chemistry74
- Ch.22 - Organic Chemistry140
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Capitolo 14, Problema 95
What mass of salt (NaCl) should you add to 1.00 L of water in an ice cream maker to make a solution that freezes at -10.0 °C? Assume complete dissociation of the NaCl and density of 1.00 g/mL for water.
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Identify the formula for freezing point depression: \( \Delta T_f = i \cdot K_f \cdot m \), where \( \Delta T_f \) is the change in freezing point, \( i \) is the van't Hoff factor, \( K_f \) is the cryoscopic constant, and \( m \) is the molality.
Determine the change in freezing point: \( \Delta T_f = 0.0 \text{ °C} - (-10.0 \text{ °C}) = 10.0 \text{ °C} \).
For NaCl, the van't Hoff factor \( i \) is 2 because it dissociates into two ions: Na\(^+\) and Cl\(^-\).
Use the known value of \( K_f \) for water, which is 1.86 \(\text{ °C kg/mol}\). Substitute the values into the formula: \( 10.0 = 2 \cdot 1.86 \cdot m \) to solve for molality \( m \).
Convert molality to mass: Use the definition of molality \( m = \frac{\text{moles of solute}}{\text{kg of solvent}} \) and the molar mass of NaCl (58.44 g/mol) to find the mass of NaCl needed.
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Freezing Point Depression
Freezing point depression is a colligative property that describes how the freezing point of a solvent decreases when a solute is added. The extent of this depression depends on the number of solute particles in the solution, not their identity. For ionic compounds like NaCl, which dissociates into two ions (Na+ and Cl-), the effect is doubled, making it crucial for calculating the required mass of solute to achieve a desired freezing point.
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Percorso guidato
01:59
Freezing Point Depression
Colligative Properties
Colligative properties are properties of solutions that depend on the ratio of solute particles to solvent molecules, rather than the type of solute. These properties include boiling point elevation, freezing point depression, vapor pressure lowering, and osmotic pressure. Understanding these properties is essential for determining how much solute to add to achieve specific physical changes in the solvent, such as lowering the freezing point of water.
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01:26Colligative Properties
Molarity and Density
Molarity is a measure of concentration defined as the number of moles of solute per liter of solution. In this context, knowing the density of water (1.00 g/mL) allows for the conversion between volume and mass, which is necessary for calculating how much NaCl to add. Since the problem involves a specific volume of water, understanding how to relate mass, volume, and molarity is key to solving the question.
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01:13Molarity
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