Lecture 3sf

Lecture 3sf - A beaker of pure water and a beaker...

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A beaker of pure water and a beaker containing a concentrated sucrose solution are placed next to each other in a covered, closed system. After a period of time, what is observed? The vapor pressure of the solution is less than the vapor pressure of pure water. P soln = x A P o A where x A < 1 The water keeps evaporating to get to its vapor pressure P o A , and since this value is > P soln , you cannot have simultaneous equilibrium between the vapor , the pure water and the solution. When the vapor pressure in the container becomes > P soln , some vapor will condense in to the solution. This results in more pure water evaporating, since the pressure in the container will be < P o A . The result is a transfer of water from the pure water into the solution. This will continue until the pure water container is empty, and all the water is transferred into the sugar solution. Now consider a related diagram: A container of sugar solution surrounded by pure water. The sugar solution is enclosed by a semipermeable membrane , meaning water, but not dissolved solute, can get through the membrane.
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What would happen after a period of time? There would be a net flow of water through the membrane from the pure water into the solution. This resultant solution would be pushed up the vertical tube. This “push” is known as osmotic pressure.
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The liquid will rise to a certain height, at which point the flow inward will equal the flow outward. The height reached is a measure of the osmotic pressure. We represent the osmotic pressure by the Greek letter pi, Π . Analysis indicates that the osmotic pressure is proportional to the molarity of the solution, and can be calculated using the relationship: Π = MRT What is the osmotic pressure of a 0.10M solution at 298 K? Π = MRT = 0.10(0.0821)(298) = 2.45 atm 1.00 atm is equivalent to 760 Torr, or 760 mm (0.76 meter) of mercury This pressure, 2.45 atm, would represent a height of mercury of 2.45(0.76 meter) = 1.86 meter. Water would go up 13.6 times higher since mercury is 13.6 times denser than water. 1.86 x 13.6 = 25.3 meters of water or about 83 feet. We thus get a substantial height of water for a 0.100 M solution. This means that we would get a measurable osmotic pressure with very small concentrations. A concentration of 0.0001M (1/1000 of .10M) would reach a height of 1/1000(25.3 meter) = 0.025 m or 2.5 cm. This is small but easily measurable, meaning that we can use osmotic pressure to detect concentrations as low as 0.0001M (equivalent to 10 -4 M or 1/10,000 M).
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A solution of 4.68 g of hemoglobin is dissolved in water to make a total volume of 125 mL. The osmotic pressure of this solution is 0.0135 atm at 300 K. Calculate the molar mass of hemoglobin. Π
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This note was uploaded on 04/03/2008 for the course CHEM 162 taught by Professor Siegal during the Spring '08 term at Rutgers.

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Lecture 3sf - A beaker of pure water and a beaker...

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