Question

A decimolar solution of potassium ferrocyanide is 50% dissociated at 300 K. Calculate the osmotic pressure of the solution.
(Given: R = 8.314 J K⁻¹ mol⁻¹)

Answer 1

Given: Molarity (C) = 0.1 mol/L

Temperature (T) = 300 K

Gas constant (R) = 8.314 J K⁻¹ mol⁻¹

Degree of dissociation (α) = 0.5

Electrolyte (K₄[Fe(CN)₆]) dissociates into 5 ions.

So, n = 5

Van’t Hoff factor (i) = 1 + α (n − 1)

= 1 + 0.5(5 − 1)

= 1 + 2

= 3

Osmotic pressure (π) = iCRT

= 3 × 0.1 × 8.314 × 300

= 3 × 0.1 × 294.2

= 748.26 J/L

Since 1 J/L = 1 Nm⁻² we get,

π = 748.28 Nm⁻²