Question

If \[0.6{\text{mL}}\] of glacial acetic acid with density \[1.06{\text{gm}}{{\text{L}}^{ - 1}}\] is dissolved in \[1{\text{Kg}}\] water and the solution froze at \[ - {0.0205^o}{\text{C}}\] . Calculate van’t Hoff factor. Given \[{{\text{K}}_{\text{f}}}\] for water is \[1.86{\text{K kgmo}}{{\text{l}}^{ - 1}}\] .

Answer 1

Hint:\[\Delta {{\text{T}}_{\text{f}}} = {{\text{K}}_{\text{f}}}{\text{m}}\] is used to determine depression in freezing point of solution where m is molality of that solution. \[{{\text{K}}_{\text{f}}}\] is the intensive property of solvent like water and does not depend upon solute or solution.

Complete step by step answer:
Depression in freezing point is the property of decrease in freezing point of the solution when some non volatile solute is dissolved with the solvent. The depression in freezing point is given by \[\Delta {{\text{T}}_{\text{f}}}\] . Freezing point is defined as the temperature at which the liquid and the solid forms of the same substance are in equilibrium and hence at this state have the same vapour pressure. As vapour pressure of the solution is less than that of pure solvent when a non volatile solute like glacial acetic acid is added to it. As freezing point is the temperature at which vapour pressure of the liquid and the solid phase are equal, therefore for the solution, this will occur at lower temperature. This is known as depression in freezing point.
Depression in freezing point can be given by \[\Delta {{\text{T}}_{\text{f}}} = {\text{T}}_{\text{f}}^o - {{\text{T}}_{\text{f}}}\] where \[{\text{T}}_{\text{f}}^o\] is freezing point of pure solvent and \[{{\text{T}}_{\text{f}}}\] is freezing point of solution. Molal depression constant is defined as the decrease in freezing point when the molality of solution is unity. Therefore, \[\Delta {{\text{T}}_{\text{f}}} = {{\text{K}}_{\text{f}}}{\text{m}}\] where \[{{\text{K}}_{\text{f}}}\] is molal depression constant and m is molality of solution.
As given in question, volume of glacial acetic acid is \[0.6{\text{mL}}\] and density is \[1.06{\text{gm}}{{\text{L}}^{ - 1}}\] , so mass of glacial acetic acid can be given by \[{\text{Mass}} = {\text{density}} \times {\text{volume}} = 1.06 \times 0.6 = 0.636{\text{g}}\] .
Moles of glacial acetic acid then can be given by \[{\text{mole}} = \dfrac{{{\text{mass}}}}{{{\text{molar mass}}}} = \dfrac{{0.636}}{{60}} = 0.0106{\text{mol}}\] .
Now using the equation: \[\Delta {{\text{T}}_{\text{f}}} = {{\text{K}}_{\text{f}}}{\text{m}}\] we can say:
 \[\Delta {{\text{T}}_{\text{f}}} = {{\text{K}}_{\text{f}}} \times {\text{molality}} = {{\text{K}}_{\text{f}}} \times \dfrac{{{\text{moles of solute}}}}{{{\text{mass of solvent}}\left( {{\text{in Kg}}} \right)}}\]
and by putting value we get:
\[\Delta {{\text{T}}_{\text{f}}} = 1.86 \times \dfrac{{0.0106}}{1} = 0.0197{\text{K}}\] .
As we know \[{\text{van't Hoff factor}} = \dfrac{{{\text{observed freezing point}}}}{{{\text{calculated freezing point}}}}\]
Therefore \[{\text{van't Hoff Factor}} = \dfrac{{0.205}}{{0.0197}} = 1.04\]

Thus, the correct answer is \[1.04\] .

Note:
Colligative properties are those properties of a solution that depends on the number of solute particles dissolved in it and do not depend on the nature of solute particles. Relative lowering of vapour pressure, osmotic pressure, elevation in boiling point and depression in freezing point are colligative properties.