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 0.046 M and 0.130 M at a temperature of 23C?

Tried using pi=MRT, but don't know what to do with the molarities given. Tried adding and subtracting the two but no correct answer.

To find the applied pressure at the end of the process, we need to calculate the osmotic pressure using the final concentrations of the solutions.

Osmotic pressure (π) can be calculated using the formula:
π = MRT

Where:
π = osmotic pressure
M = molar concentration of the solution (in mol/L)
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)

Let's calculate the osmotic pressure for both solutions:

For the first solution with a concentration of 0.046 M:
π1 = (0.046 M) x (0.0821 L·atm/(mol·K)) x (23°C + 273.15 K)
= 0.046 x 0.0821 L·atm/(mol·K) x 296.15 K
≈ 1.113 L·atm/mol

For the second solution with a concentration of 0.130 M:
π2 = (0.130 M) x (0.0821 L·atm/(mol·K)) x (23°C + 273.15 K)
= 0.130 x 0.0821 L·atm/(mol·K) x 296.15 K
≈ 3.120 L·atm/mol

Now, we need to find the difference in osmotic pressure between the two solutions:
Δπ = π2 - π1
= 3.120 L·atm/mol - 1.113 L·atm/mol
≈ 2.007 L·atm/mol

The applied pressure at the end of the process would be equal to the difference in osmotic pressure between the two solutions. Therefore, the applied pressure at the end of this process is approximately 2.007 L·atm/mol.

To determine the applied pressure at the end of the process, we can use the van 't Hoff equation, which relates the equilibrium constant (K) with the change in Gibbs free energy (∆G), temperature (T), and the gas constant (R). However, in this case, we are dealing with a semipermeable membrane and not a gas, so we need to modify the equation slightly.

The modified van 't Hoff equation for osmotic pressure is:
π = iMRT

Where:
- π is the osmotic pressure (applied pressure)
- i is the van 't Hoff factor (the number of particles the solute breaks up into when it dissolves)
- M is the molarity (concentration) of the solute
- R is the gas constant (0.0821 L·atm/(mol·K))
- T is the temperature in Kelvin

In this case, we have two containers separated by a semipermeable membrane, so the initial concentration of the waste solution is not relevant. We are interested in the final concentrations, which are 0.046 M and 0.130 M.

To solve this problem, we need to know the van 't Hoff factor (i) for the solute. The van 't Hoff factor represents how many particles the solute will dissociate or associate into when it dissolves. This value depends on the specific solute. Assuming the solute is a non-electrolyte and does not dissociate or associate in water, the van 't Hoff factor (i) would be 1.

Now that we have all the required values, we can substitute them into the equation and solve for π (the osmotic pressure):

π = (1) x (0.046 M) x (0.0821 L·atm/(mol·K)) x (273 + 23 K)

Note: We convert the temperature from Celsius to Kelvin by adding 273 to it.

Calculating the above expression, we find that the osmotic pressure (applied pressure) at the end of this process is approximately equal to 10.66 atm.

Please note that this calculation assumes ideal behavior and ignores any other factors that may affect the osmotic pressure, such as interactions between solute particles or the presence of other dissolved components.

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