Question

The osmotic pressure of 0.01 molar solution of an electrolyte is found to be 0.65 atm at 27°C. Calculate the van’t Hoff factor. What conclusion can you draw about the molecular state of the solute in the solution?

Answer 1

Given: Osmotic pressure (π) = 0.65 atm

Molarity (C) = 0.01 mol/L

Temperature (T) = 27°C = 300 K

Gas constant (R) = 0.0821 L atm mol⁻¹⋅K⁻¹

Van’t Hoff factor (i) = iCRT

`i = pi/(CRT)`

= `0.65/(0.01 xx 0.0821 xx 300)`

= `0.65/0.2463`

i = 2.64

Conclusion:

• The van’t Hoff factor (i) = 2.64, which is greater than 1, indicating that the solute is an electrolyte and undergoes dissociation in solution.
• Since ideal full dissociation of a 1 : 3 electrolyte (like AlCl₃) would give i = 4, and 1 : 2 electrolytes (like MgCl₂) give i = 3, the value 2.64 suggests partial dissociation.