Question

Define cryoscopic constant.

Answer 1

Hint: It has been observed that a solution containing non-volatile solute freezes at a temperature lower than that of the pure solvent. This decrease in the freezing point is directly proportional to the mole fraction of the solute in the solution.

Complete answer:
Freezing point is the temperature at which solid and liquid states of a substance co-exist, i.e. they have the same vapour pressure. In addition a non-volatile solute is added in a solvent as its freezing point decreases. This is called the depression in freezing point. If \[{{T}_{f}}\] and \[{{T}_{o}}\] are the freezing points of the solution containing non-volatile solute and the pure solvent, respectively. Then, the depression in freezing point (\[\Delta {{T}_{f}}\]) is given as: \[\Delta {{T}_{f}}={{T}_{o}}-{{T}_{f}}\]
\[\Delta {{T}_{f}}\] is directly proportional to the amount of non-volatile solute added per kg of the solvent. Molality is the number of moles of solute present in one kg of the solvent.
$ \Delta {{T}_{f}}\propto molality(m)$
$\Delta {{T}_{f}}={{K}_{f}}m$

$\text{or }{{K}_{f}}=\dfrac{\Delta {{T}_{f}}}{m}$

Here, \[{{K}_{f}}\] is the molal freezing point depression constant of the solvent or cryoscopic solvent. Cryoscopic constant is, therefore, defined as the ratio of the depression in freezing point to the molality of the solution. Its basic unit is \[\text{K kg mo}{{\text{l}}^{-1}}\].
If we take molality of the solution as unity, we get
$m=1$
$\Delta {{T}_{f}}={{K}_{f}}\times 1$
$\Delta {{T}_{f}}={{K}_{f}}$

We can also define a cryoscopic constant as the freezing point of the solution when one mole of the solute is dissolved in one kg of the solvent. In some cases, \[{{K}_{f}}\] may also be calculated from the thermodynamically derived equation:
\[~{{K}_{f}}=\dfrac{RT_{f}^{2}}{1000\times {{L}_{f}}}\]
where \[{{L}_{f}}\] is the latent heat of fusion per gram of the solvent.

Additional Information \[\Delta {{T}_{f}}\] of a solution is a colligative property. It depends on the mole fraction of solute and is independent of the nature of the solute. Molar mass of the solute can be determined from it. Let \[{{M}_{2}}\] be the molar mass of the solute and \[{{w}_{1}}\], \[{{w}_{2}}\] be the weights of the solute and the solvent in grams, respectively. Using the expression for molality, we get
$m=\dfrac{{{n}_{2}}}{{{w}_{1}}}$

$m=\dfrac{{{w}_{2}}\times 1000}{{{M}_{2}}\times {{w}_{1}}(g)}$
$\Delta {{T}_{f}}={{K}_{f}}m$
$\Delta {{T}_{f}}={{K}_{f}}\dfrac{{{w}_{2}}\times 1000}{{{M}_{2}}\times {{w}_{1}}(g)}$

$\text{then, }{{M}_{2}}=\dfrac{{{K}_{f}}\times {{w}_{2}}\times 1000}{\Delta {{T}_{f}}\times {{w}_{1}}(g)}$

Note: Cryoscopic constant has a specific value for any given solvent. Physical quantities without appropriate units have no significance. So, never write \[{{K}_{f}}\] without units and use proper symbols for different physical quantities.