2. A benzene (B)-toluene (T) solution is in equilibrium with its vapor at 80°C. If уB = 0.670, calculate XB and the total vapor pressure of the solution. The vapor pressures at 80°C of benzene and toluene are 100.4 kPa and 38.7 kPa, respectively. (answer: XB = 0.439, P = 65.8 kPa)**

2. A benzene (B)-toluene (T) solution is in equilibrium with its vapor at 80°C. If уB = 0.670, calculate XB and the total vapor pressure of the solution. The vapor pressures at 80°C of benzene and toluene are 100.4 kPa and 38.7 kPa, respectively. (answer: XB = 0.439, P = 65.8 kPa)**

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter11: Properties Of Solutions

Section: Chapter Questions

Problem 70E: The vapor pressures of several solutions of water-propanol (CH3CH2CH2OH) were determined at various...

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Concept explainers

Solutions

Solutions can be considered as homogeneous mixtures which contain two or more than two chemical substances. The chemical substance which is the greater part is defined as solvent. And the chemical substance that is dissolved in a solvent is referred to as solute. Solution chemistry plays an important role in chemistry since it supports many commercial applications of solutions.

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2. A benzene (B)-toluene (T) solution is in equilibrium with its vapor at 80°C. If уB -0.670, calculate XB and
the total vapor pressure of the solution. The vapor pressures at 80°C of benzene and toluene are 100.4 kPa
and 38.7 kPa, respectively. (answer: XB-0.439, P= 65.8 kPa)**
Calculate the equilibrium concentration (represented as a mole fraction) of carbon dioxide in fat given that

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A solution is prepared by dissolving 40.00 g of NaCl (f.w. = 58.44 g mol–1), a non-volatile solute, in enough water (m.w. = 18.02 g mol–1) to result in exactly 1 L of solution at 25 °C. Assume the density of the solution is that of pure water (dsolution = 1.000 g mL–1). The ebullioscopic constant (Kb) for water is 0.513 °C m–1. The cryoscopic constant (Kf) for water is 1.86 °C m–1. The vapor pressure of pure water is 0.0313 atm. Find the osmotic pressure in atm to three decimal places
A solution is prepared by dissolving 40.00 g of NaCl (f.w. = 58.44 g mol–1), a non-volatile solute, in enough water (m.w. = 18.02 g mol–1) to result in exactly 1 L of solution at 25 °C. Assume the density of the solution is that of pure water (dsolution = 1.000 g mL–1). The ebullioscopic constant (Kb) for water is 0.513 °C m–1. The cryoscopic constant (Kf) for water is 1.86 °C m–1. The vapor pressure of pure water is 0.0313 atm. Determine the boiling point of the solution(in C to 2 decimal places)
A solution is prepared by dissolving 40.00 g of NaCl (f.w. = 58.44 g mol–1), a non-volatile solute, in enough water (m.w. = 18.02 g mol–1) to result in exactly 1 L of solution at 25 °C. Assume the density of the solution is that of pure water (dsolution = 1.000 g mL–1). The ebullioscopic constant (Kb) for water is 0.513 °C m–1. The cryoscopic constant (Kf) for water is 1.86 °C m–1. The vapor pressure of pure water is 0.0313 atm. Determine the following: Boiling point of solution (in °C to two decimal places) Freezing point of solution (in °C to two decimal places) Vapor pressure of the solution (in atm to three decimal places) Osmotic pressure (in atm to three decimal places)