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 0.163 m in sodium chloride and 0.019 m in magnesium sulfate? assume complete dissociation for electrolytes.

The minimum pressure should be 901.79 kPa

Explanation:

Step 1: Data given

Temperature = 25°C

Molarity of sodium chloride = 0.163 M

Molarity of magnesium sulfate = 0.019 M

Step 2: 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

the answer is 75.0%.