Question

0.6 ml of glacial acetic acid with density $ 1.06{\text{g/mol}} $ is dissolved in 1 kg water and the solution froze at $ - {0.0205^0}C $ . Calculate Van’t Hoff factor:
 $ {K_f} $ for water is $ 1.86{\text{KKgmo}}{{\text{l}}^{{\text{ - 1}}}} $ .

Answer 1

Hint: We need to find the mass of acetic acid. Further we need to apply the Van’t Hoff formula which we shall get from the colligative property of depression in freezing point. We shall use the formula below and substitute the given values to calculate the answer.

Formula used: $ \Delta {{\text{T}}_f}{\text{ = }}{{\text{K}}_f} \times n $
Here $ \Delta {T_f} $ is the freezing point
 $ n $ is the number of moles.
Also, $ {\text{Vant Hoff Factor = }}\dfrac{{{\text{observed freezing point}}}}{{{\text{calculated freezing point}}}} $

Complete step by step solution:
We already know the density of acetic acid.
And the volume of acetic acid is 0.6ml.
Mass of acetic acid $ = 0.6{\text{ml}} \times 1.06{\text{gm}}{{\text{l}}^{{\text{ - 1}}}} = 0.636{\text{g}} $
No of moles of acetic acid $ = \dfrac{{0.636}}{{60}} = 0.0106{\text{mol}} $
Using the equation:
 $ \Delta {{\text{T}}_f}{\text{ = }}{{\text{K}}_f} \times n $
 $ \Delta {T_f} = 1.86 \times 0.0106 = 0.0197{\text{K}} $
Now,
 $ {\text{Vant Hoff Factor = }}\dfrac{{{\text{observed freezing point}}}}{{{\text{calculated freezing point}}}} $
 $ {\text{Vant Hoff Factor = }}\dfrac{{0.0205}}{{0.0197}} = 1.04 $
So, the Van’t Hoff Factor is $ 1.04. $

Note:
We should keep in mind that the formula for determining the van't Hoff factor is measured value/calculated value. The van't Hoff factor can be applied to any of the colligative properties. For our work with colligative properties, we will always use the ideal value for the solution. Colligative properties are important properties of solutions as they describe how the properties of the solvent will change as solute (or solutes) is (are) added. It means that the property, in this case change in temperature, depends on the number of particles dissolved into the solvent and not the nature of those particles. The van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.
For example, we can assume sucrose. Sucrose does not break into particles in solution. It's Van't Hoff factor is 1. The theoretical van't Hoff factor of $ AlC{l_3} $ is 4, because in theory it ought to break up into 4 ions when dissolved in water. In reality it is 3.