Question

Please help explain how to obtain the (i) in the van't Hoff factor. My text states that it is a ratio of the experimentally measured value of a colligative property to the value expected if the solute did not dissocite into ions when it dissoloved. (I don't understand that definition) Also the text provided a few examples: the value for "i" for NaCl is 2, MgSO₄ is 2, CaCl₂ is 3 and 4 for Na₃PO_​4​. (I don't understand how to obtain that "i" value. Please help! Many Thanks!

Answer 1

Vant hoff factor is used for non ideal solution. Ideal solution is considered as non electrolyte means non dissociable. Like benzene in toluene is non dissociable so it is ideal solution. But an aqueous NaCl solution means it is dissociable. NaCl dissociates in water as Na+ and Cl- ion. So the colligative properties such as osmotic pressure, elevation of boiling point all this properties will show double of its expected values because in stead of one particle, there is two particles for each molecule. Here Vant hoff factor is 2. CaCl₂ also similarly dissociates into Ca²⁺ and two Cl- ions means total 3 ions. So all the colligative properties will show triple values than expected. Here Vant hoff factor is 3.

So to obtain the Vant hoff factor of any substance first determine whether it is dissociable or not. If not dissociable then its 1 and will be called ideal solution but if dissociable, then find out how many ions are formed from each molecule. The number of ions formed will be the Vant hoff factor.