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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 �C when reverse osmosis stops? 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 79.1 L of fresh water at 20 �C with an applied pressure of 50.0 bar.

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IthaloAbreu

Answer:

Part A. 2.05 M

Part B. 170.7 L

Explanation:

Part A

For an ideal fluid, we can use the ideal gas law to find out the final molar concentration (C = n/V):

PV = nRT

Where P is the pressure, V is the volume, R is the gas constant (0.08314 bar.L/mol.K), n is the number of moles, and T is the temperature. Thus,

C = P/RT

For T = 20°C + 273 = 293 K

C = 50/(0.08314*293)

C = 2.05 M

Part B

The initial concentration is 1.1 M, so for 1 L of seawater, the volume (V) produced of fresh water is:

Ci/(1 - V) = C

1.1/(1 - V) = 2.05

1.1 = 2.05 - 2.05V

2.05V = 0.95

V = 0.463 L

Thus

1 L ------------ 0.463 L

x   ------------- 79.1 L

By a simple direct three rule:

0.463x = 79.1

x = 170.7 L

Part- A The final concentration of the seawater at [tex]20^oC[/tex] when reverse osmosis stops is 2.05M.

Part-B 170.7 liters of seawater are needed to produce 79.1 L of fresh water at  [tex]20^oC[/tex]  with an applied pressure of 50.0 bar.

Solving for each part on by one:

  • Part A

Ideal fluid:

For an ideal fluid, we can use the ideal gas law to find out the final molar concentration [tex](C = n/V)[/tex] :

Ideal gas law states that:

[tex]PV = nRT[/tex]

where

  • P is the pressure,
  • V is the volume,
  • R is the gas constant (0.08314 bar.L/mol.K),
  • n is the number of moles, and
  • T is the temperature.

For T = 20°C + 273 = 293 K

[tex]C =\frac{P}{RT}\\\\C = \frac{50}{0.08314*293}\\\\C = 2.05 M[/tex]

  • Part B

The initial concentration is 1.1 M, so for 1 L of seawater, the volume (V) produced of fresh water is:

[tex]\frac{C_i}{1-V} = C\\\\\frac{1.1}{1-V}= 2.05\\\\1.1 = 2.05 - 2.05V\\\\2.05V = 0.95\\\\V = 0.463 L[/tex]

Thus

1 L  ⇒ 0.463 L

x   ⇒   79.1 L

Now, solving for x:

[tex]0.463x = 79.1\\\\x = 170.7 L[/tex]

Find more information about Ideal gas law here:

brainly.com/question/12873752

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