Ch 13 HW, Part 2 Colligative Properties

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Feb 20, 2024

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Ch. 13 HW, Part 2: Colligative Properties Ch. 13 HW, Part 2: Colligative Properties Properties of Solutions This is Part 2 for the Chapter 13 Homework which includes problems that cover colligative properties. 1/1 point A 0.200 m solution of which of the following solutes will have the lowest vapor pressure? O Glucose (a molecular compound) () Licl () KClI u O Aicl, () caCl, Feedback General Feedback Vapor-pressure lowering is a colligative property that depends on the total concentration of non-volatile solute particles in the solution, and not on the kind or identity of the solute. Determine which of the solutes will result in the highest concentration of solute particles. Remember that soluble ionic compounds will break apart into their ions when dissolved in water, and molecular compounds will not.

1/1 point A solution is prepared by adding 40.00 g of lactose (molar mass 342.3 g/mol) to 110.0 g of water at 55 °C. The vapor pressure of pure water at 55 °C is 118.0. Calculate the partial pressure of the water above the solution in torr. Answer Parameters: ¢ Report your answer with THREE significant figures. o IF NEGATIVE, a - sign must be placed IN FRONT of the number. o The number must be in decimal form and not include any other units or symbols B Feedback General Feedback You will use Raoult's Law to calculate the partial pressure of the water above the solution, Psgution- First, calculate the moles of lactose and the moles of water to find the total moles. Keep four significant figures to avoid rounding errors. Next, calculate the mole fraction of water, X,y,ter = moles water / total moles. Now, use Raoult's Law to find Psojution = Xwater) (PCwater) where PO ater = 118.0 torr. | 1F4.270 g of sucrose (C15H27011) are dissolved in 15.20 g of water, what will be the boiling point of the resulting solution? (K, for water = 0.5120°C/m) Answer Parameters: + Report your answer with FOUR significant figures. * IF NEGATIVE, a - sign must be placed IN FRONT of the number. * The number must be in decimal form and not include any other units or symbols 100.4202

1/1 point Calculate the freezing point of a solution containing 20.0 g of KCl and 2200.0 g of water. The molal freezing point depression constant for water is 1.86°C/m. Answer Parameters: * Report your answer with THREE significant figures. e |FNEGATIVE, a - sign must be placed IN FRONT of the number. * The number must be in decimal form and not include any other units or symbols oass 1/ 1 point A 2.05 m aqueous solution of an unknown solute has a boiling point of 102.1°C. The boiling point elevation constant, Ky, for water is 0.512 °C/m. Which one of the following could be the unknown compound? Feedback General Feedback The molality, new boiling point and Ky, are given. You can calculate the change in boiling point, AT}, by subtracting the boiling point of pure water from the boiling point given for the solution given in the problem. Now solve ATy, = i Ky, m for i and determine which of the substances given would have a van't Hoff factor equal to . n 1/1 point Which of the following aqueous solutions is expected to have the lowest freezing point? ) 0.010mNazPO, ~) 0.040 m glycerin (C3HgO3) () 0.020mKBr © Allof the solutions would be expected to have the same freezing point. h 1/ 1 point Fill in the blanks \/ As the concentration of a non-volatile solute in a solution increases, the freezing point , the boiling point ,and the vapor pressure of Nl decreases the solution Feedback General Feedback Freezing point, boiling point and vapor pressure are colligative properties that depend on the concentration of the solute particles but not on the kind or identity of the solute particles. The effect on these properties is a lowering of the freezing point, an elevation of the boiling point, and lowering of the vapor pressure.

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