Question

The osmotic pressure of a 0.0103 molar solution of an electrolyte is found to be 0.70 atm at 27°C. Calculate the van't Hoff factor. (R = 0.082 L atm mol⁻¹ K⁻¹). What conclusion do you draw about the molecular state of the solute in the solution?

Answer 1

Given: Molarity (C) = 0.0103 mol/L

Osmotic pressure (π) = 0.70 atm

Temperature (T) = 27°C = 300 K

Gas constant (R) = 0.082 L atm mol⁻¹ K⁻¹

π = iCRT

i = `pi/(CRT)`

i = `0.70/(0.0103 xx 0.082 xx 300)`

i = `0.70/0.25338`

i = 2.76

Since i = 2.77 (much greater than 1), it indicates that the solute is electrolytic and undergoes significant dissociation in solution, likely forming 3 particles per formula unit (e.g., a salt like AlCl₃ or BaCl₂).