Concept explainers
Interpretation: The solubility of benzoic acid in water and benzene with measured freezing point depression as equal, less or greater than 5.12°C needs to be determined if the solubility of benzoic acid in water is 0.34 g/100 mL and 10.0 g/100 mL in benzene at 25°C.
Concept Introduction: Colligative properties are the properties of solution which depend on the number of particles present in the solution. Some common examples of colligative properties are depression in freezing point, elevation in boiling point, osmotic pressure, lowering in vapor pressure etc. The number of particles is shown with the help of van’t Hoff factor ‘i’.
The mathematical expression for depression in freezing point and elevation in boiling point are as given below:
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Chapter 17 Solutions
Chemical Principles
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- The vapor pressures of several solutions of water-propanol (CH3CH2CH2OH) were determined at various compositions, with the following data collected at 45C: H2O Vapor pressure(torr) 0 74.0 0.15 77.3 0.37 80.2 0.54 81.6 0.69 80.6 0.83 78.2 1.00 71.9 a. Are solutions of water and propanol ideal? Explain. b. Predict the sign of Hsoln for water-propanol solutions. c. Are the interactive forces between propanol and water molecules weaker than, stronger than, or equal to the interactive forces between the pure substances? Explain. d. Which of the solutions in the data would have the lowest normal boiling point?arrow_forwardWhat would be the freezing point of a solution formed by adding 1.0 mole of glucose (a molecular compound) to the following amounts of water? a. 250 g (0.25 kg) b. 500 g (0.500 kg) c. 1000 g (1.000 kg) d. 2000 g (2.000 kg)arrow_forwardThe organic salt [(C4H9)4N][ClO4] consists of the ions (C4H9)4N+ and ClO4. The salt dissolves in chloroform. What mass (in grams) of the salt must have been dissolved if the boiling point of a solution of the salt in 25.0 g chloroform is 63.20 C? The normal boiling point of chloroform is 61.70 C and Kb = 3.63 C kg mol1. Assume that the salt dissociates completely into its ions in solution.arrow_forward
- The freezing-point depression of a 0.091-m solution of CsCl is 0.320C. The freezing-point depression of a 0.091-m solution of CaCl2 is 0.440C. In which solution does ion association appear to be greater? Explain.arrow_forwardThe freezing point of 0.109 m aqueous formic acid is 0.210C. Formic acid, HCHO2, is partially dissociated according to the equation HCHO2(aq)H+(aq)+CHO2(aq) Calculate the percentage of HCHO2 molecules that are dissociated, assuming the equation for the freezing-point depression holds for the total concentration of molecules and ions in the solution.arrow_forwardA 0.109 mol/kg aqueous solution of formic acid, HCOOH, freezes at −0.210 °C. Calculate the percent dissociation of formic acid.arrow_forward
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