(a)
Interpretation:
Freezing point and boiling point of the solutions to be calculated.
5g
Concept introduction:
Freezing point: Freezing point is defined as the temperature at which liquid into solid.
Boiling point: Boiling point is temperature at which the liquid into gas.
Elevation of boiling point: Elevation of boiling point of a liquid is occurrence that boiling point of a liquid would be greater when another substance is added to it. Resulting, the boiling point becomes higher than the pure solvent.
Depression of freezing point: Depression of freezing point is reducing in freezing point of a solvent when added non volatile solute. Volatile substance means easily vaporized at room temperature and non volatile means does not easily evaporate at room temperature.
(b)
Interpretation:
Freezing point and boiling point of the solutions to be calculated.
2.0g
Concept introduction:
Freezing point: Freezing point is defined as the temperature at which liquid into solid.
Boiling point: Boiling point is temperature at which the liquid into gas.
Elevation of boiling point: Elevation of boiling point of a liquid is occurrence that boiling point of a liquid would be greater when another substance is added to it. Resulting, the boiling point becomes higher than the pure solvent.
Depression of freezing point: Depression of freezing point is reducing in freezing point of a solvent when added non volatile solute. Volatile substance means easily vaporized at room temperature and non volatile means does not easily evaporate at room temperature.
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Chapter 11 Solutions
Chemistry
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- An aqueous solution contains 0.180 g of an unknown, nonionic solute in 50.0 g of water. The solution freezes at 0.040 C. What is the molar mass of the solute?arrow_forwardEqual numbers of moles of two soluble, substances, substance A and substance B, are placed into separate 1.0-L samples of water. a The water samples are cooled. Sample A freezes at 0.50C, and Sample B freezes at l.00C. Explain how the solutions can have different freezing points. b You pour 500 mL of the solution containing substance B into a different beaker. How would the freezing point of this 500-mL portion of solution B compare to the freezing point of the 1.0-L sample of solution A? c Calculate the molality of the solutions of A and B. Assume that i = 1 for substance A. d If you were to add an additional 1.0 kg of water to solution B, what would be the new freezing point of the solution? Try to write an answer to this question without using a mathematical formula. e What concentration (molality) of substances A and B would result in both solutions having a freezing point of 0.25C? f Compare the boiling points, vapor pressure, and osmotic pressure of the original solutions of A and B. Dont perform the calculations; just state which is the greater in each ease.arrow_forwardA solution was prepared by dissolving 0.800 g of sulfur, Sg, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and boiling point of the solution.arrow_forward
- Vodka is advertised to be 80 proof. That means that the ethanol (C2H5OH) concentration is 40% (two significant figures) by volume. Assuming the density of the solution to be 1.0 g/mL, what is the freezing point of vodka? The density of ethanol is 0.789 g/mL.arrow_forwardCalculate the freezing point of 525 g of water that contains 25.0 g of NaCl. Assume i, the vant Hoff factor, is 1.85 for NaCl.arrow_forward
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