Domanda del libro di testo
List the following aqueous solutions in order of increasing boiling point: 0.120 m glucose, 0.050 m LiBr, 0.050 m Zn(NO3)2.
1410
views
Ch.13 - Properties of Solutions
Brown14th EditionChemistry: The Central ScienceISBN: 9780134414232
Non è quello che usi tu?Cambia libro di testoScegli un libro di testo diverso
TOC.tableOfContents
- Ch.1 - Introduction: Matter, Energy, and Measurement177
- Ch.2 - Atoms, Molecules, and Ions231
- Ch.3 - Chemical Reactions and Reaction Stoichiometry216
- Ch.4 - Reactions in Aqueous Solution189
- Ch.5 - Thermochemistry175
- Ch.6 - Electronic Structure of Atoms168
- Ch.7 - Periodic Properties of the Elements150
- Ch.8 - Basic Concepts of Chemical Bonding177
- Ch.9 - Molecular Geometry and Bonding Theories200
- Ch.10 - Gases179
- Ch.11 - Liquids and Intermolecular Forces113
- Ch.12 - Solids and Modern Materials154
- Ch.13 - Properties of Solutions157
- Ch.14 - Chemical Kinetics199
- Ch.15 - Chemical Equilibrium128
- Ch.16 - Acid-Base Equilibria164
- Ch.17 - Additional Aspects of Aqueous Equilibria163
- Ch.18 - Chemistry of the Environment105
- Ch.19 - Chemical Thermodynamics164
- Ch.20 - Electrochemistry171
- Ch.21 - Nuclear Chemistry122
- Ch.22 - Chemistry of the Nonmetals82
- Ch.23 - Transition Metals and Coordination Chemistry79
- Ch.24 - The Chemistry of Life: Organic and Biological Chemistry16
Brown14th EditionChemistry: The Central ScienceISBN: 9780134414232Non è quello che usi tu?Cambia libro di testo
Capitolo 13, Problema 72
List the following aqueous solutions in order of decreasing freezing point: 0.040 m glycerin (C3H8O3), 0.020 m KBr, 0.030 m phenol (C6H5OH).
Guida verificata passo dopo passo1
Step 1: Understand that the freezing point of a solution is affected by the number of solute particles in the solution, which is described by the colligative property known as freezing point depression.
Step 2: Recall 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 freezing point depression constant, and \( m \) is the molality of the solution.
Step 3: Determine the van't Hoff factor \( i \) for each solute: Glycerin (C3H8O3) is a non-electrolyte, so \( i = 1 \); KBr is an electrolyte that dissociates into K+ and Br-, so \( i = 2 \); Phenol (C6H5OH) is a weak electrolyte, but for simplicity, assume \( i = 1 \).
Step 4: Calculate the effective molality for each solution by multiplying the molality by the van't Hoff factor: Glycerin: \( 0.040 \times 1 = 0.040 \), KBr: \( 0.020 \times 2 = 0.040 \), Phenol: \( 0.030 \times 1 = 0.030 \).
Step 5: Rank the solutions in order of decreasing freezing point. The solution with the lowest effective molality will have the highest freezing point. Therefore, the order is: Phenol (0.030 m), Glycerin (0.040 m), KBr (0.040 m).
Concetti chiave
Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.
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 solution rather than their identity. This means that solutions with more dissolved particles will have lower freezing points.
Video consigliato:
Percorso guidato
01:59
Freezing Point Depression
Van't Hoff Factor (i)
The Van't Hoff factor (i) indicates the number of particles into which a solute dissociates in solution. For example, KBr dissociates into two ions (K+ and Br-), giving it an i value of 2, while non-electrolytes like glycerin and phenol do not dissociate, so their i values are 1. This factor is crucial for calculating the effective concentration of solute particles affecting freezing point.
Video consigliato:
Percorso guidato
02:50Van't Hoff Factor
Molality (m)
Molality 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 because it directly relates to the number of solute particles in a given mass of solvent. In this question, the molalities of the solutions will help determine their respective freezing point depressions.
Video consigliato:
Percorso guidato
00:40Molality
Pratica correlata
Domanda del libro di testo
(a) Does a 0.10 m aqueous solution of NaCl have a higher
boiling point, a lower boiling point, or the same boiling
point as a 0.10 m aqueous solution of C6H12O6?
1109
views
Domanda del libro di testo
Using data from Table 13.3, calculate the freezing and boiling points of each of the following solutions: (a) 0.22 m glycerol (C3H8O3) in ethanol, (b) 0.240 mol of naphthalene (C10H8) in 2.45 mol of chloroform, (c) 1.50 g NaCl in 0.250 kg of water, (d) 2.04 g KBr and 4.82 g of glucose (C6H12O6) in 188 g of water.
1130
views
Domanda del libro di testo
At 20 °C, the vapor pressure of benzene (C6H6) is 75 torr, and that of toluene (C7H8) is 22 torr. Assume that benzene and toluene form an ideal solution. (b) What is the mole fraction of benzene in the vapor above the solution described in part (a)?
2271
views
1
comments
Domanda del libro di testo
Using data from Table 13.3, calculate the freezing and boiling points of each of the following solutions: (a) 0.25 m glucose in ethanol; (b) 20.0 g of decane, C10H22, in 50.0 g CHCl3; (c) 3.50 g NaOH in 175 g of water, (d) 0.45 mol ethylene glycol and 0.15 mol KBr in 150 g H2O.
1576
views