Skip to main content
Ch.13 - Solutions & Their Properties

Chapter 13, Problem 115

A solution prepared by dissolving 3.00 g of ascorbic acid (vitamin C, C6H8O6) in 50.0 g of acetic acid has a freez- ing point that is depressed by ΔT = 1.33 °C below that ofpure acetic acid. What is the value of the molal freezing- point-depression constant for acetic acid?

Verified Solution
Video duration:
2m
This video solution was recommended by our tutors as helpful for the problem above.
458
views
Was this helpful?

Video transcript

Hello. In this problem we are told when 9.5 g of lithium is dissolved in 550 g of benzene resulting solution freezes at 5500.30 degrees Celsius lower than that appear. Benzene grass to calculate the value of the freezing point constant for benzene. So we will make use in of the equation that describes the depression of the freezing point. For a solution is given by. Then the change in the freezing point is equal to the van Hoff factor times the freezing point constant times the morality. So in this problem we have with the it is insoluble in benzene. So that means it's van Hoff factor will be one. Our change in the freezing point then is equal 2.30°C. And the other thing that we need then is the morality which recall is defined as the moles of our salute which is the lithium straight per kilogram of our Sullivan which is benzene. So we'll put what we know about the lithium straight in the numerator. So we have 9.5 g. I'm with the um stay right In 550 grams of benzene. So we'll convert our mass of lithium lithium steroid using a smaller mass. one world lithium spirit has a mass of two or .47 g. And then we will convert our massive benzene from grams to kilograms. So our grams of lithium Sterrett cancels and our grams of benzene cancels we're left with in moles of lithium steer a per kilogram of bending. So this is a morality. It's equal to the 0.5946. Making use of our equation. Then we will solve for the freezing point constant, it's equal to then the change in the freezing point divided by van Hoff factor, divided by morality. Our change of freezing point is 0.0.0.30 degree Celsius. Van Hoff factor is one And our morality is .05946 allow. So this works out then to 5.0°C per morality. This is then the freezing point constant for Benzie. Thanks for watching. Hope this helps.
Related Practice
Textbook Question
What is the vapor pressure in mm Hg of the following solu-tions, each of which contains a nonvolatile solute? The vapor pressure of water at 45.0 °C is 71.93 mm Hg. (b) A solution of 10.0 g of LiCl in 150.0 g of water at 45.0 °C, assuming complete dissociation
1300
views
Textbook Question
Cyclopentane 1C5H102 and cyclohexane 1C6H122 are vola- tile, nonpolar hydrocarbons. At 30.0 °C, the vapor pres- sure of cyclopentane is 385 mm Hg, and the vapor pressure of cyclohexane is 122 mm Hg. What is Xpentane in a mixture of C5H10 and C6H12 that has a vapor pressure of 212 mm Hg at 30.0 °C?
332
views
Textbook Question
A solution prepared by dissolving 5.00 g of aspirin, C9H8O4, in 215 g of chloroform has a normal boiling point that is elevated by ΔT = 0.47 °C over that of pure chloro- form. What is the value of the molal boiling-point-elevation constant for chloroform?
491
views
Textbook Question
A solution of citric acid, C6H8O7, in 50.0 g of acetic acid has a boiling point elevation of ΔT = 1.76 °C. What is the molality of the solution if the molal boilin# g-point-elevation constant for acetic acid is Kb = 3.07 1°C kg2>mol.
797
views
Textbook Question
What osmotic pressure in mm Hg would you expect for an aqueous solution of 11.5 mg of insulin 1mol. weight = 59902 in 6.60 mL of solution at 298 K? What would the height of the water column be in meters? The density of mercury is 13.534 g/mL at 298 K.
749
views
Textbook Question
Human blood gives rise to an osmotic pressure of approxi-mately 7.7 atm at body temperature, 37.0 °C. What must the molarity of an intravenous glucose solution be to give rise to the same osmotic pressure as blood?
966
views