Part A The wastewater solution from a factory containing high levels of salts ne
ID: 990998 • Letter: P
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 0.059 M and 0.190 M at a temperature of 23 C? Express your answer with the appropriate units.
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 1.68×105 torr . Calculate the maximum molar concentration of a solution that can be purified by reverse osmosis using this membrane at 25 C. Express your answer with the appropriate units.
Explanation / Answer
(a) Applied pressure would be the difference of osmotic pressure() of two different concentration solutions.
We know that, = CRT
where R is gas constant and its value is 0.0821 (atm *L)/(mol* K)
T = 23 +273 = 296 K
C for the solution in first container = 0.059M and C for the solution in second container = 0.190 M
for first container, = 0.059 mol L-1 x 0.0821 atm x L x mol-1 x K-1 x 296 K = 1.43 atm
for second container, = 0.190 mol L-1 x 0.0821 atm x L x mol-1 x K-1 x 296 K = 4.62 atm
Since the applied pressure would be the difference of these two pressures,
So, 4.62 atm - 1.43 atm = 3.19 atm (Answer)
(b)
Given, = 1.68 x 105 torr, we need convert pressure in atm.
So, 1.68 x 105 torr x 1 atm/760 torr = 221 atm
T = 25 + 273 = 298 K
= CRT
C = /RT = 221 atm/(0.0821 atm x L x mol-1 x K-1 x 298 K)
C = 9.03 mol L-1 or 9.03 M (Answer)