Question

A 0.99 m aqueous solution of an ionic compound with the formula mx has a freezing point of -2.6 ∘c . calculate the van't hoff factor (i) for mx at this concentration.

Answer 1

Formula for the depression in freezing point is:

$\Delta T_f = i\times k_f\times m$  -(1)

where $\Delta T_f$ is depression in freezing point,

$i$ is Van't Hoff factor,

$k_f$ is molal freezing point depression constant, and

$m$ is molality of the solution.

Molality of the solution, $m = 0.99 m$  (given)

Molal freezing point depression constant of water, $k_f = 1.86^{o}C/m$

Depression in freezing point of solution, $\Delta T_f = T_{water solvent} - T_{solution}$

$\Delta T_f = {0^{o}C} - ({-2.6^{o}C}) = 2.6^{o}C$

Substituting the values in equation (1):

$\2.6^{o}C = i\times 1.86^{o}C/m\times 0.99 m$

$i = \frac{2.6^{o}C}{1.86^{o}C/m\times 0.99 m} = 1.412$

Hence, the Van't Hoff factor (i) for $HX$ is 1.412.