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2 years ago

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What is the minimum pressure in kPa that must be applied at 25 °C to obtain pure water by reverse osmosis from water that is 0.1

63 M in sodium chloride and 0.019 M in magnesium sulfate? Assume complete dissociation for electrolytes.

1 answer:

Yuri [45]2 years ago

6 0

Answer:

The minimum pressure should be 901.79 kPa

Explanation:

<u>Step 1: </u>Data given

Temperature = 25°C

Molarity of sodium chloride = 0.163 M

Molarity of magnesium sulfate = 0.019 M

<u>Step 2:</u> Calculate osmotic pressure

The formula for the osmotic pressure =

Π=MRT.

⇒ with M = the total molarity of all of the particles in the solution.

⇒ R = gas constant = 0.08206 L*atm/K*mol

⇒ T = the temperature = 25 °C = 298 K

NaCl→ Na+ + Cl-

MgSO4 → Mg^2+ + SO4^2-

M = 2(0.163) + 2(0.019 M)

M = 0.364 M

Π = (0.364 M)(0.08206 atm-L/mol-K)(25 + 273 K)

Π = 8.90 atm

(8.90 atm)(101.325 kPa/atm) = 901.79 kPa

The minimum pressure should be 901.79 kPa

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$w=-873.6J$

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