Question

Calculate the Freezing point of a molar aqueous solution of $ {\text{KCl}} $ . (Density of the solution = $ {\text{1}}{\text{.04 g }}{{\text{L}}^{{\text{ - 1}}}} $ , $ {{\text{K}}_{\text{f}}}{\text{ = 1}}{\text{.86 K Kg mo}}{{\text{l}}^{{\text{ - 1}}}} $ )

Answer 1

Hint: Addition of potassium chloride to water to make a solution, depresses the freezing point of the solvent and this depression in freezing point is a colligative property that is dependent on the number of moles of the solute. We shall calculate the mass of the compound using the density and then the moles and thus, the molarity of the compound. Then, we shall calculate the freezing point using the formula given.
Formula Used: $ \Delta {{\text{T}}_{\text{f}}} = {\text{i}} \times {{\text{K}}_{\text{f}}} \times {\text{m}} $
Where “i” is the Van’t Hoff factor, $ {{\text{K}}_{\text{f}}} $ is the cryoscopic constant, and m is the molality of the solution.

Complete step by step solution:
For the $ {\text{KCl}} $ solution, the particles dissociate to form two ions, the potassium chloride dissociates to form one potassium cation and one chloride anion. Hence the Van’t Hoff Factor is equal to 2.
Therefore the depression in the freezing point is,
  $ \Delta {{\text{T}}_{\text{f}}} = 2 \times {\text{1}}{\text{.86}} \times {\text{m}} $
As per the question, the Density of the solution = $ {\text{1}}{\text{.04 g }}{{\text{L}}^{{\text{ - 1}}}} $
Density of the solution = $ \dfrac{{{\text{mass of the solution}}}}{{{\text{volume of the solution}}}} $
Therefore the mass of the solution = $ 1.04 \times 1000 = 1040 $ grams.
The molarity of the solution = 1 $ {\text{mol/ litre}} $ = 1 mole of potassium chloride per litre of the solution.
The molecular weight of potassium chloride = $ 39 + 35.5 = 74.5 $ gram per mole
Therefore one mole of potassium chloride = $ 74.5 $ grams, this amount of potassium chloride is present in the solution.
The mass of water in the solution = $ {\text{1040 - 74}}{\text{.5 = 965}}{\text{.5 grams}} $ = $ 0.9655 $ Kilograms
The molality of the solution = $ \dfrac{{{\text{moles of the solute}}}}{{{\text{mass of the solvent in Kg}}}} $ = $ \dfrac{{\text{1}}}{{{\text{0}}{\text{.9655}}}}{\text{ = 1}}{\text{.035}}\left( {\text{m}} \right) $
Putting the value in the equation for the depression of freezing point we get,
 $ \Delta {{\text{T}}_{\text{f}}} = 2 \times {\text{1}}{\text{.86}} \times {\text{1}}{\text{.035 = 3}}{\text{.85}} $
Therefore the reduced freezing point of water = $ 273 - 3.85 = 269.15 $ Kelvin.

Note:
The properties that are not dependent upon the nature of the solute present in the solution, but only on the number of the solute particles or the moles of the solute particles are called the “colligative properties”. There are four colligative properties overall and they are: the lowering of vapour pressure, the depression of freezing point, the elevation in the boiling point, and the osmotic pressure.