The boiling point of water is 100.0°C at 1 atmosphere. How many grams of potassium chloride (74.55 g/mol), must be dissolved in 275.0 grams of water to raise the boiling point by 0.450°C? Refer to the table for the necessary boiling or freezing point constant. Solvent Formula Kp (°C/m) Kf (°C/m) Water H20 0.512 1.86 Ethanol CH3CH2OH 1.22 1.99 Chloroform CHC13 3.67 Benzene C6H6 2.53 5.12 Diethyl ether CH3CH2OCH2CH3 2.02 g potassium chloride.

Explain why it is the freezing point depression of a solvent does not depend on the identity of the solute. Explain why it is the freezing point depression of a solvent does depend on the concentration of the solute.

Boiling Point Elevation/Freezing Point Depression T = m K where, for freezing point depression: T = T(pure solvent) - T(solution) and for boiling point elevation: T = T(solution) - T(pure solvent) m = (# moles solute / Kg solvent) Kb = boiling point elevation constant. Kf = freezing point depression constant. Kb and Kf depend only on the SOLVENT. Below are some common values. Use these values for the calculations that follow. Solvent Formula Kb(°C / m) Kf(°C / m) Water H2O 0.512 1.86 Ethanol CH3CH2OH 1.22 1.99 Chloroform CHCl3 3.67   Benzene C6H6 2.53 5.12 Diethyl ether CH3CH2OCH2CH3 2.02

Boiling Point ElevationThe boiling point of a solution INCREASES with the number of dissolved solute particles. This is summarized by the expression: Tb = m i KbWhere: Tb is the increase in the boiling point = Tb(solution) - Tb(pure solvent) m is the molality of the solution = (# moles solute / kg solvent) i is the "van't Hoff" factor = (# moles of solute particles / mole of solute)   For nonelectrolytes: i = 1   For strong electrolytes: i = number of (cations + anions)   For weak electrolytes: 1 < i < number of (cations + anions) Kb is the boiling point elevation constant. It depends only on the SOLVENT. Example:Na2SO4 is a soluble salt. It breaks into two Na+ cations and one SO42- anion in solution.i = 2 + 1 = 3 CH3COOH is a weak acid. It breaks into a few CH3COO - anions and a few H+ cations. 1 < i < 2.

You have a sucrose solution of 28°Brix or 28 grams of sugar per 100 grams solution in a ripe grape ready to be made into wine. What is the molality of the solution? Look up the freezing point depression constant KF and figure out what the new freezing temperature would be for that solution.

The boiling point of an aqueous solution is 101.45 Celsius. What is the freezing point?

Name Time M T W R F The freezing point of pure benzene (CH,) is 5.50°C. The freezing point of a solution containing 5.00 g of an unknown non-dissociating solute in 444 g of benzene is 5.07°C. K, of benzene = 4.90°C/m. Show complete calculations for each question below. 1. What is the freezing point depression for this solution?

Consider the sugar–water phase diagram of Figure 9.1. Initially, saturated liquid solution 900 g of water at 60°C is cooled to 20°C (68°F), some of the sugar precipitates as a solid. How much of the solid sugar in kg will come out of solution upon cooling to 20°C?

An experiment was conducted to determine the effect of glucose on the freezing point of water. Experimental data showed thatthe freezing point depression of water in solution was –2.6°C when 1.0 g of glucose was dissolved in 10g of solvent. Calculatethe expected freezing point for such solution and compare the expected freezing point to the value found experimentally. Give aplausible explanation for any discrepancies. (M.W. of glucose= 180.16g/mol; i=1). Show all your work.

Raoult’s Law states: the vapor pressure (Psolution) of a solution is proportional to the mole fraction of the solvent present in the solution. Psolution = Xsolvent P°solvent Here,  P°solvent ­is the vapor pressure of pure solvent. Calculate the vapor pressure of a solution which is prepared by dissolving 1.00 mole of glycerol (nonvolatile) and 4.00 moles of water at 25oC. Given: the vapor pressure of water is 25.0 torr at 25oC. Group of answer choices 5.00 torr 20.0 torr 6.25 torr 15.0 torr