Domanda del libro di testo
A glucose solution contains 55.8 g of glucose (C6H12O6) in 455 g of water. Determine the freezing point and boiling point of the solution.
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Ch.14 - Solutions
Tro5th EditionChemistry: A Molecular ApproachISBN: 9780134874371
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TOC.tableOfContents
- Ch.1 - Matter, Measurement & Problem Solving117
- Ch.2 - Atoms & Elements106
- Ch.3 - Molecules and Compounds128
- Ch.4 - Chemical Reactions and Chemical Quantities56
- Ch.5 - Introduction to Solutions and Aqueous Solutions84
- Ch.6 - Gases112
- Ch.7 - Thermochemistry96
- Ch.8 - The Quantum-Mechanical Model of the Atom60
- Ch.9 - Periodic Properties of the Elements86
- Ch.10 - Chemical Bonding I: The Lewis Model91
- Ch.11 - Chemical Bonding II: Molecular Shapes, VSEPR & MO Theory85
- Ch.12 - Liquids, Solids & Intermolecular Forces51
- Ch.13 - Solids & Modern Materials48
- Ch.14 - Solutions93
- Ch.15 - Chemical Kinetics114
- Ch.16 - Chemical Equilibrium68
- Ch.17 - Acids and Bases137
- Ch.18 - Aqueous Ionic Equilibrium154
- Ch.19 - Free Energy & Thermodynamics91
- Ch.20 - Electrochemistry105
- Ch.21 - Radioactivity & Nuclear Chemistry74
- Ch.22 - Organic Chemistry141
Tro5th EditionChemistry: A Molecular ApproachISBN: 9780134874371Non è quello che usi tu?Cambia libro di testo
Capitolo 14, Problema 82
Calculate the molar mass of an unknown molecular (nonelectrolyte) compound in an aqueous solution containing 35.9 g of the compound in 150.0 g of water with a freezing point of -1.3 °C.
Guida verificata passo dopo passo1
Identify the freezing point depression formula: \( \Delta T_f = i \cdot K_f \cdot m \), where \( i \) is the van't Hoff factor (1 for nonelectrolytes), \( K_f \) is the freezing point depression constant for water (1.86 °C kg/mol), and \( m \) is the molality.
Calculate the freezing point depression \( \Delta T_f \) by subtracting the observed freezing point from the normal freezing point of water (0 °C): \( \Delta T_f = 0 - (-1.3) = 1.3 \) °C.
Rearrange the formula to solve for molality \( m \): \( m = \frac{\Delta T_f}{i \cdot K_f} \). Substitute \( \Delta T_f = 1.3 \) °C, \( i = 1 \), and \( K_f = 1.86 \) °C kg/mol.
Calculate the molality \( m \) using the values from the previous step.
Use the definition of molality \( m = \frac{\text{moles of solute}}{\text{kg of solvent}} \) to find the moles of solute, then use the mass of the solute to calculate the molar mass: \( \text{Molar mass} = \frac{\text{mass of solute (g)}}{\text{moles of solute}} \).
Concetti chiave
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Molar Mass
Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of all atoms in a molecule. Understanding molar mass is essential for converting between grams and moles, which is crucial for stoichiometric calculations in chemistry.
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02:11
Molar Mass Concept
Freezing Point Depression
Freezing point depression is a colligative property that describes the lowering of a solvent's freezing point when a solute is added. The extent of freezing point depression is directly proportional to the number of solute particles in the solution, which can be calculated using the formula ΔTf = i * Kf * m, where ΔTf is the change in freezing point, Kf is the freezing point depression constant, and m is the molality of the solution.
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01:59Freezing Point Depression
Molality
Molality (m) is a measure of concentration defined as the number of moles of solute per kilogram of solvent. It is particularly useful in colligative property calculations, such as freezing point depression, because it accounts for the mass of the solvent rather than the total volume of the solution, providing a more accurate representation of the solution's properties.
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00:40Molality