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# The freezing point of 0.0262mol fraction of acetic acid in benzene is 277.4K. acetic acid exists partly as a dimer. Calculate the equilibrium constant for dimerization.(approximately)(freezing point of ${{C}_{6}}{{H}_{6}}=278.4k$ and ${{K}_{f}}of=5Kkg/mol$)(a)2.6(b)3.2(c)4.3(d)5.8

Last updated date: 20th Sep 2024
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Hint: Freezing point refers to the temperature at which a liquid becomes a solid, as with the melting point, increased pressure usually raises the freezing point. The freezing point is lower than the melting point in the case of mixtures and for certain organic compounds like fats.

Colligative properties are properties that vary in terms of solvent concentration, but not in terms of the form of solution. For the example above, what this implies is that people in colder climates do not actually need salt to have the same impact on the roads-any solvent will work. The higher the solute concentration, however, the more these properties change.
Elevation of the boiling point is the increase of the boiling point of a solvent due to the addition of a solvent. Similarly, freezing point depression is a loss of the freezing point of a liquid due to the addition of a solution. Indeed, when a solvent's boiling point increases, its freezing point shrinks. An example of that would be to apply salt to an icy sidewalk. The solvent (salt) decreases the Ice freezing point, causing the ice to melt at a lower temperature.
Given:
$\Delta {{H}_{fusion}}=10.042KJ/mol$
${{T}_{f}}=278.4k$
$2C{{H}_{3}}COOH{{\left( C{{H}_{3}}COO \right)}_{2}}$
$C\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ O$
$C\left( 1-\alpha \right)\ \ \ \ \ \ \ \ C\dfrac{\alpha }{2}$
${{k}_{c}}=\dfrac{C\dfrac{\alpha }{2}}{{{C}^{2}}{{\left( 1-\alpha \right)}^{2}}}=\dfrac{C\alpha }{2{{C}^{2}}{{\left( 1-\alpha \right)}^{2}}}$

$\Delta {{T}_{f}}=i\times {{K}_{f}}\times m$
So, Dimer is forming : $\eta =2$

$\alpha =\left( 1-i \right)\times 2$

Putting value in equation:
$\Delta {{T}_{f}}=i\times {{K}_{f}}\times m$
$\Delta {{T}_{f}}=1-\dfrac{\alpha }{2}\times {{K}_{f}}\times m$
$\Delta {{T}_{f}}=1-\dfrac{\alpha }{2}\times 5\times m$

$m=\dfrac{0.02\times 1000}{0.98\times 78}=0.262\,molal$
So, $C=0.262\,molal$
$\Delta {{T}_{f}}=5\times 0.262\times \left( 1-\dfrac{\alpha }{2} \right)$
$278.4-277.4=5\times 0.262\times \left( 1-\dfrac{\alpha }{2} \right)$
$\alpha =0.48$
${{k}_{c}}=\dfrac{C\alpha }{2{{C}^{2}}{{\left( 1-\alpha \right)}^{2}}}$
${{k}_{c}}=\dfrac{0.262\times 0.48}{2\times {{\left( 0.262 \right)}^{2}}{{\left( 1-0.48 \right)}^{2}}}$
${{k}_{c}}=3.39\sim 3.2$

Note:
-So, $\Delta {{T}_{f}}=T_{f}^{0}-{{T}_{f}}$
$\alpha$= Degree of dissociation
i= Van't Hoff factor
$\eta$= No. of ions obtained on association or dissociation
- Boiling point and freezing points are colligative properties of a solution.
-The factor of van' t Hoff I is a measure of the impact of a solution on colligative properties such as osmotic pressure, relative decrease in vapour pressure, of the elevation in boiling point and the depression in freezing point.