Question

Given the following mixture of two compounds 40.00 mL of X (MW =67.00 g/mol)(density 0.906 g/mL) and 690.00 mL of Y (81.00 g/mol))(density 0.858 g/mL). If R = 0.08206 L.atm/mol/K, calculate the osmotic pressure of the solution at 26 degrees C.

Answer 1

Osmotic pressure – P

P = CRT = nRT/ V where n = number of moles and V = total volume of solution

P = (n1 +n2) RT/V, 

Density = mass/volume

Compound X, mass = 40 x 0.906 = 36.24g

Number of moles of X = 36.24/67 = 0.5409

Compound Y, mass = 690 x 0.858 = 592.02g

Number of moles of Y= 592.02/81 = 7.309

Total volume of the solution = 40 + 690 = 730ml = 0.730L

R = 0.08206 L.atm/mol/K

Temperature = 26⁰C = 299k

Therefore, P = (n1 + n2) RT/V = (0.5409 + 7.309) x 0.08206 x 299/0.730

P = 263.8420atm.