P soln B P water = x water P 0 water P 0 water = x water x water = P soln B P water = mol H 2 O mol H 2 O + mol NaBr = 1000 18 1000 18 + 1 = 55 . 5 56 . 5 028 1.0 points The following diagram shows a solution on the left (dark shade) and just the solvent on the right (light shade) separated by a semiperme- able membrane. Which of the following diagrams best rep- resents the final state of this system after equilibrium is achieved? 1. 2.
To (st22362) – Homework 2 – Fakhreddine – (52420) 10 3. 4. correct 5. Explanation: Only the solvent goes through the mem- brane thus increasing the volume on the solu- tion side and decreasing volume on the solvent side. The solution side is therefore diluted slightly and the shade is lightened somewhat. The solute itself (the darker color) cannot pass through the membrane so the right side MUST also stay the same color of pure solvent (the lighest shade shown). 029 1.0 points Polyacrylamide is a water-soluble polymer whose aqueous solution containing 25 g/L de- velops an osmotic pressure of 0.54 torr at 25 ◦ C. Find the approximate molecular weight of the polymer sample. 1. 1133 g/mol 2. 35,000 g/mol 3. 860,000 g/mol correct 4. 350,000 g/mol 5. 8,600,000 g/mol Explanation: π = 0.54 torr density = 25 g/L T = 25 ◦ C + 273 = 298 K Here we can use the equation π = M R T for osmotic pressure, to our advantage. How- ever, to use the “standard” value of R , we’ll have to convert our pressure from torr to atm: 0.54 torr parenleftBig 1 atm 760 torr parenrightBig = 0.00071 atm Then use the equation for osmotic pressure: 0 . 00071 atm = M parenleftbigg 0 . 08206 L · atm K · mol parenrightbigg (298 K) = 2 . 9 × 10 − 5 mol / L However, we know the solution is 25 g/L, so we can determine the molecular weight of the polymer with a simple division: 25 g / L 2 . 9 × 10 − 5 mol / L = 860,000 g/mol 030 1.0 points Water from a local stream is added to one side of the U-tube shown below. Pure water is placed in the tube on the other side of the semipermeable membrane. With the left side open to barometric pressure of 1.0 atm and 1.5 atm applied to the right side, the two liquids do not move. 1.00 atm 1.15 atm B A What is the effective molarity ( i.e. , molarity of “particles” such as cations, anions, etc. ) responsible for this osmotic pressure from the natural water from the local stream? You can assume the van’t Hoff factor is one. 1. 0.03 molar 2. 6 × 10 − 5 molar 3. 0.006 molar correct 4. 0.2 molar
To (st22362) – Homework 2 – Fakhreddine – (52420) 11 Explanation: 031 1.0 points Solution A contains 0.5 grams of solute A and solution B contains 0.5 grams of solute B (both A and B are nonelectrolytes). Other than the solutes, the solutions are identical (volume, temperature, etc .). Now you mea- sure the osmotic pressure of each solution and find that the osmotic pressure of solution B is twice that of solution A. What is the relation- ship between the molecular weights of solutes A and B? 1. The molecular weight of solute B is four times that of solute A. 2. The molecular weight of solute A is four times that of solute B. 3. The molecular weight ratio cannot be determined from this experiment. 4. The molecular weight of solute B is twice that of solute A. 5. The molecular weight of solute A is twice that of solute B. correct Explanation: Because the solution with B has twice the osmotic pressure, it must also have twice the number of dissolved particles (number of moles of B). The only way to have more moles for B (knowing the same mass of A and B was used) is that B has a smaller molecular weight than A. For the osmotic pressure to be ex-