Van’t Hoff introduced a factor i, known as the van’t Hoff factor, to account for the extent of dissociation or association. This factor i is defined as:
$i = \frac {\text{Normal molar mass}}{\text{Abnormal molar mass}}$
$i=\frac {\text{Observed colligative property}}{\text{Calculated colligative property}}$
$i = \frac {\text {Total number of moles of particles after association/dissociation}}{\text {Number of moles of particles before association/dissociation}}$
Inclusion of van’t Hoff factor modifies the equations for colligative properties as follows:
Relative lowering of vapour pressure of solvent,
$\frac {p_1^0 - p_1}{p_1^o}= i \frac {n_2}{n_1}$
Elevation of Boiling point, $\Delta T_b = i K_b m$
Depression of Freezing point, $\Delta T_f = i K_f m$
Osmotic pressure of solution, $\prod = i n_2 R T / V$
Solved Example
Example -1
2 g of benzoic acid (C6H5COOH) dissolved in 25 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant for benzene is 4.9 K kg/mol. What is the percentage association of acid if it forms dimer in solution? Solution
The given quantities are: $W_B = 2$ g, $K_f= 4.9 $K kg mol–1; $W_A = 25 $g,
Now $\Delta T_f= 1.62 $K
Now
$\Delta T_f= K_f \times \frac { (W_B \times 1000)}{(M_B \times W_A )}$
or $M_B = 241.98$ g/mol
Thus, experimental molar mass of benzoic acid in benzene is
= 241.98 g/mol
Now consider the following equilibrium for the acid:
$2 C_6H_5COOH -> (C_6H_5COOH)^2$
If x represents the degree of association of the solute then we would have (1 – x ) mol of benzoic acid left in unassociated form and correspondingly x/2 as associated moles of benzoic acid at equilibrium.
Therefore, total number of moles of particles at equilibrium is:
1-x + x/2=1 -x/2
Thus, total number of moles of particles at equilibrium equals van’t Hoff factor i.
Now
$i = \frac {\text{Normal molar mass}}{\text{Abnormal molar mass}}$
$1 -x/2 = \frac {122}{241.98} $
or x =.992
