In reverse osmosis, water flows out of a salt solution until the osmotic pressure of the solution equals the applied pressure. If a pressure of 47.0 bar is applied to seawater, what will be the final concentration of the seawater at 20 °C when reverse osmosis stops?

Mc=?

Assuming that seawater has a total ion concentration (a.k.a colligative molarity) of 1.10 Mc, calculate how many liters of seawater are needed to produce 64.8 L of fresh water at 20 °C with an applied pressure of 47.0 bar.
L=?

To find the final concentration of the seawater when reverse osmosis stops, we will use the formula:

Mc = (π * Vw) / (RT * MW)

Where:
Mc is the final concentration of the seawater
π is the applied pressure in atmospheres (1 bar = 1 atm)
Vw is the volume of water permeated in liters
R is the ideal gas constant (0.0821 L * atm / mol * K)
T is the temperature in Kelvin
MW is the molecular weight of water (18.015 g/mol)

First, let's convert the pressure from bar to atm:
1 bar = 0.9869 atm (approximately)

So, the applied pressure is 47.0 bar * 0.9869 atm/bar = 46.3653 atm

Now, let's convert the temperature from Celsius to Kelvin:
20 °C + 273.15 = 293.15 K

We are given that the total ion concentration of seawater is 1.10 Mc.

Now, we can rearrange the formula and solve for Mc:

Mc = (π * Vw) / (RT * MW)

Substituting the known values:
1.10 Mc = (46.3653 atm * Vw) / (0.0821 L * atm / mol * K * 293.15 K * 18.015 g/mol)

Simplifying, we get:
1.10 Mc = (46.3653 atm * Vw) / (4770.6756 L * atm / mol * g)

To find the volume of seawater required to produce 64.8 L of fresh water, we'll use the formula:

L = Vw + Vf

Where:
L is the total volume,
Vw is the volume of water permeated, and
Vf is the volume of fresh water produced.

Given that Vf = 64.8 L, we can rearrange the formula and solve for Vw:

Vw = L - Vf

Substituting the values:
Vw = (unknown) - 64.8 L

Since we do not have the value for L, we cannot calculate the specific volume of seawater required.

To solve these questions, we need to use the equation for osmotic pressure and the concentration of the seawater.

The osmotic pressure (Π) can be calculated using the equation:

Π = McRT

Where:
- Mc is the molar concentration of the solution in mol/L
- R is the ideal gas constant (0.0831 L bar/mol K)
- T is the temperature in Kelvin (20 °C = 293 K)

From the question, we know that the applied pressure is 47.0 bar. At equilibrium, the osmotic pressure will be equal to the applied pressure (Π = 47.0 bar).

Plugging in the values, we can solve for the molar concentration (Mc):

47.0 bar = Mc * 0.0831 L bar/mol K * 293 K

Simplifying the equation gives:

Mc = 47.0 bar / (0.0831 L bar/mol K * 293 K)

Mc ≈ 2.07 mol/L

So, the final concentration of the seawater when reverse osmosis stops is approximately 2.07 mol/L.

Now, let's move on to the second question:

To calculate the number of liters of seawater needed to produce 64.8 L of fresh water, we'll set up a proportion using the molar concentration and the volume ratios.

2.07 mol/L seawater = x mol/L fresh water

The ratio of seawater to fresh water will be the same as the ratio of the volumes when the moles of both are equal. Therefore, we have:

1.10 Mc seawater / x L fresh water = V seawater / 64.8 L fresh water

Rearranging the equation, we get:

V seawater = 1.10 Mc seawater * (64.8 L fresh water / x)

Substituting the known values, we can solve for V seawater:

V seawater ≈ 1.10 mol/L * (64.8 L / 2.07 mol/L)

V seawater ≈ 34.54 L

Therefore, approximately 34.54 liters of seawater are needed to produce 64.8 liters of fresh water at 20 °C with an applied pressure of 47.0 bar.