## van’t Hoff factor

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

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