e following relation describes the volume change in a water (solvent – species 1) and salt (solute – species 2) solution: V=1.001+16.625n2 +56.092 n23/2 +119.4n22 where V is expressed in units of liters, and n2 in kmole. (1) Obtain an expression for partial V2 in terms of n2, and determine partial V2 when x2 is approching 0 in liter per kmole. (2) Obtain an expression for partial V1 one in terms of n2, and one in terms of x1

e following relation describes the volume change in a water (solvent – species 1) and salt (solute – species 2) solution: V=1.001+16.625n2 +56.092 n23/2 +119.4n22 where V is expressed in units of liters, and n2 in kmole. (1) Obtain an expression for partial V2 in terms of n2, and determine partial V2 when x2 is approching 0 in liter per kmole. (2) Obtain an expression for partial V1 one in terms of n2, and one in terms of x1

Chemistry: Principles and Practice

3rd Edition

ISBN:9780534420123

Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Chapter12: Solutions

Section: Chapter Questions

Problem 12.72QE

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The following relation describes the volume change in a water (solvent – species 1) and salt (solute – species 2) solution: V=1.001+16.625n2 +56.092 n23/2 +119.4n22 where V is expressed in units of liters, and n2 in kmole. (1) Obtain an expression for partial V2 in terms of n2, and determine partial V2 when x2 is approching 0 in liter per kmole. (2) Obtain an expression for partial V1 one in terms of n2, and one in terms of x1.
Phenol and water form non-ideal liquid mixtures. When 7.32 g of phenol and 7.95 g of water are mixed together at 60 °C they form two immiscible liquid phases with mole fractions of phenol of 0. 0 42 and 0 .161 . (a) Calculate the overall mole fraction of phenol in the mixture. (b) Use the lever rule to determine the relative amounts of the two phases .
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Show that (demonstrate) for an ideal binary solution, where 1 is the solvent and 2 is the solute, the increase in the boiling point of the solvent (ΔT) due to the presence of the solute is given by the relation: ∆T=Kebb2=(T'eb-Teb) where Keb=[R(Teb)2M1]/1000∆vapH1] Keb=ebulloscopic constant Please show all of the steps and calculations used to demonstrate the question. As well as a brief explanation of the principle used.
You are handed a 15.85% by mass aqueous KCl (m.m. = 74.5513 g mol–1) 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: Vapor pressure of the solution in atm
You are handed a 15.85% by mass aqueous KCl (m.m. = 74.5513 g mol–1) 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 one decimal place) 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)
You are handed a 15.85% by mass aqueous KCl (m.m. = 74.5513 g mol–1) 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 in degrees Celcius
You are handed a 15.85% by mass aqueous KCl (m.m. = 74.5513 g mol–1) 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: Osmotic Pressure in atm
You are handed a 15.85% by mass aqueous KCl (m.m. = 74.5513 g mol–1) 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: Freezing Point of Solution in Degrees Celcius
Can you please answer number 6 and show all of the steps to the solution