Question

Determine the osmotic pressure (in psi) , at 75.0^o F, of an aqueous solution that is 2.60 % sodium sulfate by mass.

Density _solution = 1.09 g/mL

Answer 1

Sodium sulfate (Na₂SO₄) is an electrolyte

hence

Osmotic pressure () = iMRT.

M = molarity

i = vant hoff factor

Given , concentration of sodium sulfate solution = 2.60 %

Hence in 100 g solution 2.60 g Sodium sulfate is present .

Density of the solution= 1.09 g/ml

Hence volume of the solution = (100/1.09) = 91.74 mL

Molar mass of sodium sulfate = 142 g/mol.

Hence , molar concentration = [ (2.60*1000)/(142*91.74)]

= 0.19958 mol/L

1 mol Na_{2​​​​​​}SO₄ dissociates completely to give 2 mol Na⁺ and 1 mol SO₄^2- . Hence i = 3.

T = 75.00 F = [ (75-32)*5/9] + 273.15 = 297.039 K

R = 0.082 L-atm/mol.K

Hence, = (3*0.19958 *0.082*297.039) = 14.584 atm.

Now, 1 atm = 14.69 psi

Then 14.584 atm = (14.584*14.696) = 214.326 psi.