In reverse osmosis, water flows out of a salt solution until the osmotic pressur

Question

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 50.0 bar is applied to seawater, what will be the final concentration of the seawater at 20 DegreeC when reverse osmosis stops? Number M_c 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 43.3 L of fresh water at 20 DegreeC with an applied pressure of 50.0 bar. Number L

Explanation / Answer

We first convert the applied pressure in bars to atmosphere by noting that 1 atm = 1.01325 bars.

Therefore, 50 bar = (50 bar/1.01325 bar)*1 atm = 49.346 atm.

When the flow of water stops, the osmotic pressure = applied pressure.

We know that osmotic pressure, is related to the molar mass of the solute as

= MRT where M is the molar mass of the solute, R is the gas constant = 0.082 L-atm/mol.K and T is the absolute temperature = (273 + 20) = 293K

Therefore, 49.346 atm = M.(0.082 L-atm/mol.K)(293 K) = M.(24.026 L-atm/mol)

or, (49.346/24.026) mol/L = M ===è M = 2.054 mol/L = 2.05 Mc

The molar concentration of the salt is 2.05 Mc (ans).

Now, for the second part, we start with a molar concentration of 1.10 Mc of the dissolved salts and end up with a molar concentration of 2.05 Mc. We shall apply the dilution equation as

V1 x M1 = V2 x M2 where 1 and 2 denote the initial and final states, V denotes volume and M denotes molar concentration. Let V1 = V and V2 = (V – 43.3) (since we pull out 43.3 L fresh water from salt water and the total volume of salt water is assumed to be V)

Hence, V x 1.10 Mc = (V – 43.3) x 2.05 Mc

or, V = ( V – 43.3) x 1.864

or, V = 1.864 – 80.7112

or, 0.864V = 80.7112 ===è V = 93.4157

The volume of sea water needed is 93.41 L (correct to 2 decimal places) (ans)

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