A solution prepared by dissolving 3.00 g of ascorbic acid (vitami...

Question

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?

Accepted Answer

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.

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?

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