Question

A) 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 66.0 bar is applied to seawater, what will be the final concentration of the seawater at 20 °C when reverse osmosis stops?

B) 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 86.2 L of fresh water at 20 °C with an applied pressure of 66.0 bar.

Answer 1

1 bar= 0.989 atm 66 bar= 66*0.989=65.274 atm

osmotic pressure = iCRT, C =concentration i= number of ions Na+ and Cl- =2

T= 20 deg.c= 20+273.15= 293.15K and R= 0.08206 L.atm/mole.K

                  C= Osmotic pressure/ iRT= 65.274/2*0.08206*293.15=1.356 moles/L

Ithe intital water= 1.10 Mc and final water= 1.356*2=2.712 ( Mc= i*M)

MV1= M2V2

V1= (M2V2/M1) V2= volume of water to be produced= 86.2L

V1- water removed= 86.2

V1= (M2/M1)* (V1-86.2)

V1= (2.712/1.10)*(V1-86.2)

V1= 2.5*(V1-86.2)

1.5V1= 2.5*86.2

V1= 2.5*86.2/1.5=143.7 L of sea water.

B)