A solution of a water soluble polymer having a molar mass of 3.20*10^5g/mol is p

Question

A solution of a water soluble polymer having a molar mass of 3.20*10^5g/mol is placed in an osmometer and allowed to equilibrate. After equilibration at 298K, the difference in height between the top of the solution in the osmometer's tube and the water level outside the tube is found to be 8.00cm. The total volume of the polymer solution is 10mL. What mass of this polymer was initially added to the osmometer? Pleases assume that the density of the aqueous polymer solution remains 1.00g/mL. Please answer this with detailed step, thank you

Explanation / Answer

The difference in height is the pressure difference between two sides.

80 mm/760 mm = 0.105 atm

Using the form of the van't Hoff equation PV = nRT, the number of moles of polystyrene is
n = (0.105 atm)(0.01L) / (0.0821 L atm mol-1 K-1)(298 K) = 4.29 x 10-5 mol

A solution of a water soluble polymer having a molar mass of 3.20*105g/mol

Mass of polymer is 4.29 x 10-5 mol x 3.20*105g/mol = 13.73 g of polymer

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