Part A The wastewater solution from a factory containing high levels of salts ne

Question

Part A

The wastewater solution from a factory containing high levels of salts needs to be diluted before it can be released into the environment. Therefore, two containers of waste solution are separated by a semipermeable membrane and pressure is applied to one container, forcing only water molecules through the membrane and diluting the waste solution in the other container. As dilution continues, higher and higher pressures are needed to counteract the natural tendency for the water molecules to have a net flow back toward the more concentrated solution.
What was the applied pressure at the end of this process if the final concentrations of the solutions were 0.051 M and 0.190 M at a temperature of 23 degreesC?

Part B

There are limitations to the amount of pressure that can be applied during reverse osmosis because of the physical nature of the membrane. A membrane has been developed that can withstand osmotic pressures, , of up to 1.63×105 torr .
Calculate the maximum molar concentration of a solution that can be purified by reverse osmosis using this membrane at 25 degreesC.

Explanation / Answer

At a constant temprature osmotic pressure is directly propotional to the concentration so

Osmotic pressure = CRT

C = Molar concentration

concentration difference = 0.190-- 0.051 =0.139

osmotic pressure = 0.139 × 0.082×296= 3.373 atm

PART B

Pressure = 1.63 ×105 Torr

= 1.63 × 105 ÷ 760 = 0.225 atm

0.225 = C × 0.082 × 298

C = 9.2 ×10--3 M

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