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=?
I got 28.26 for Mc....can someone confirm this answer and help me find the liters for the next part.
To determine the final concentration of the seawater at 20°C when reverse osmosis stops, we need to consider the osmotic pressure and the applied pressure. Given that the applied pressure is 47.0 bar, we can use this information to find the final concentration of the seawater.
The osmotic pressure (Π) of a solution can be calculated using the formula:
Π = i * n * R * T
Where:
i = Van't Hoff factor (the number of particles per molecule or formula unit in solution)
n = concentration of solute (in moles per liter)
R = ideal gas constant (0.0821 L * atm / (mol * K))
T = temperature in Kelvin (20°C + 273.15 = 293.15 K)
In reverse osmosis, the applied pressure counteracts the osmotic pressure, so we can set the osmotic pressure equal to the applied pressure:
47.0 bar = (i * n * R * T)
Rearranging the equation, we can solve for the concentration (n):
n = (47.0 bar) / (i * R * T)
As we're given the total ion concentration of the seawater (1.10 Mc), which includes all the dissolved ions, we assume the Van't Hoff factor (i) is equal to 2. This is because seawater generally contains a mix of ions such as sodium chloride (NaCl), which dissociates into two particles: Na+ and Cl-.
Now, substituting the values into the equation, we can calculate the final concentration (n) of the seawater:
n = ((47.0 bar) / (2 * 0.0821 L * atm / (mol * K) * 293.15 K))
n ≈ 0.954 Mc
So, the final concentration of the seawater when reverse osmosis stops is approximately 0.954 Mc.
Now, let's calculate the number of liters of seawater needed to produce 64.8 L of fresh water at 20°C with an applied pressure of 47.0 bar.
To do so, we can use the formula for the volume of a solution (V) in terms of moles:
V = (mass of solute) / (molarity)
We can rearrange the formula to solve for the mass of solute:
mass of solute = (V) * (molarity)
Taking the known values:
V = 64.8 L
molarity (Mc) = 1.10 Mc
mass of solute = (64.8 L) * (1.10 Mc)
mass of solute ≈ 71.28 mol
Given that seawater has an approximate density of 1.0 g/mL, we can convert the mass of the solute to volume using the equation:
Volume = (mass of solute) / (density)
Converting the mass to grams:
mass of solute ≈ 71.28 mol * (1 mol/L) * (1000 g/1 mol)
mass of solute ≈ 71,280 g
Now, we can calculate the volume:
Volume = (71,280 g) / (1.0 g/mL)
Volume ≈ 71,280 mL
Finally, converting the volume to liters:
Volume ≈ 71,280 mL * (1 L/1000 mL)
Volume ≈ 71.28 L
Therefore, approximately 71.28 liters of seawater are needed to produce 64.8 liters of fresh water at 20 °C with an applied pressure of 47.0 bar.
To find the final concentration of the seawater at 20 °C when reverse osmosis stops, we need to use the osmotic pressure equation:
π = i * Mc * R * T
Where:
π is the osmotic pressure
i is the van't Hoff factor (the number of particles dissolved in the solution)
Mc is the concentration of the solute in mol/L
R is the ideal gas constant (0.0821 L·atm/(mol·K) or 0.0831 L·bar/(mol·K))
T is the temperature in Kelvin
Since seawater is a solution of various salts and minerals, we can assume an average value of the total ion concentration as 1.10 Mc.
At 20 °C, T would be 20 + 273.15 = 293.15 K.
The applied pressure is given as 47.0 bar.
Now, we can rearrange the osmotic pressure equation to solve for Mc:
Mc = π / (i * R * T)
Substituting the given values:
Mc = 47.0 bar / (1 * 0.0831 L·bar/(mol·K) * 293.15 K)
Calculating this expression gives the final concentration of the seawater when reverse osmosis stops.
To calculate the number of liters of seawater needed to produce 64.8 L of fresh water at 20 °C with an applied pressure of 47.0 bar, we can use the formula:
L = (Vs * Mc * Vw) / (P * ΔC)
Where:
L is the number of liters of seawater needed
Vs is the volume of the solution (seawater) in liters
Mc is the initial concentration of the solute (seawater) in mol/L
Vw is the volume of fresh water produced in liters
P is the applied pressure in bar
ΔC is the change in concentration of the solute
In this case, the initial concentration of seawater is 1.10 Mc. The volume of fresh water produced is 64.8 L, and the applied pressure is 47.0 bar.
To calculate the change in concentration (ΔC), we subtract the final concentration from the initial concentration:
ΔC = Mc - final concentration
Once we have all the values, we can substitute them into the formula to find the number of liters of seawater needed to produce 64.8 L of fresh water at 20 °C with an applied pressure of 47.0 bar.