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 48.0 bar is applied to seawater, what will be the final concentration of the seawater at 20

Answer 1

Posmotic = iMRT

where..
i = van't hoff factor = # ions 1 unit of solute dissociates into = 2 (if the salt is NaCl or KCl.. 1 NaCl --> 1Na + 1Cl-... 1 --> 2 ions.. right? so i=2)
M = molarity = the variable you want
R = gas constant = 0.08206 Latm/molK
T = temp in K = 20 + 273.15 = 293.15K

you know...
Posmotic = Papplied = 44.0bar x (1atm / 1.01325bar) = 43.42 atm

so... rearrange and solve...
M = Posmotic / (i x R x T)
M = 43.42atm x (1 / 2) x (1 molK / 0.08206 Latm) x (1 / 293.15K) = 0.902 mol / L = 0.902 M

hint.. enter this in your calculator
43.42 / 2 / 0.08206 / 293.15 =
and round to 3 sig figs

*** Part 2 ***

the Mc in part 2 is literally "i x M"

In part 1 we said i = 2 and M = 0.902M
so.. in part 1, "i x M" = 1.804

In other words, we start with a solution of Mc = 1.10 and end with solution of Mc = 1.804 by removing water. How much water do we remove?

from a mole balance of the solute.. the "ions" in solute....
M1V1 = M2V2

rearranging
V1 = V2 x (M2 / M1)

Since V1 - water removed = V2 .. think about it, we start with V1 and remove some water to get V2 right? substituting..
V1 = (V1 - water removed) x (M2 / M1)

rearranging..
V1 = V1 x (M2 / M1) - water removed x (M2 / M1)

rearranging
(V1 - V1 x M2 / M1) = - water removed x (M2 / M1)

factoring and rearranging
V1 x (M2 / M1 - 1) = +water removed x (M2 / M1)

and finally
V1 = water removed x (M2 / M1) / (M2 / M1 - 1)

and if you really want to, you could simplify that right hand side a bit more
(M2 / M1) / (M2 / M1 - 1) = 1 / (1 - 1/(M2/M1)) = 1 / (1 - M1/M2)

anyway.. solving..
V1 = 35.5L fresh H2O / (1 - (1.10Mc / 1.804Mc)) = 90.7gal salt water