
Calculate depression of freezing point for $ 0.56 $ molal aq. solution of $ KCl $ .
(Given: $ {K_f}({H_2}O)\, = 1.8\,Kkg/mol $ ).
Answer
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Hint : Depression in freezing point is a colligative property of a solution. Colligative properties are those properties that are dependent on the ratio of the number of moles of the solute to the number of moles of solvent in the solution.
Complete Step-by-step Solution
The colligative properties of the solution do not depend on the chemical nature of the solution. In terms of solution, these properties tell us how the properties of a solution are connected to the solute concentration in the given solution. There are $ 4 $ colligative properties that are exhibited by the solutions. These are:
-Elevation in Boiling Point
-Depression in Freezing Point
-Relative Lowering in the Vapour Pressure
-Osmotic Pressure
Now, let’s talk about Depression in Freezing Point. According to Raoult’s law when you mix a non-volatile solute in a solvent it lowers the vapour pressure and becomes equal to the vapour pressure of the solvent at a lower temperature. This difference between the freezing point of pure solvent and that of a solution is called the depression in freezing point. It is given by the expression;
$ \Delta {T_f} = i \times {K_f} \times m $
Where, $ i = $ Van’t hoff factor
$ {K_f} = $ cryoscopic constant and $ m = $ molality of the solution
Now, we are given $ m = 0.56 $ , $ {K_f} = 1.8 $ . We will calculate the value of $ i $ and will put all the three values in the formula to find the value of $ \Delta {T_f} $ . As $ KCl $ is a strong acid it will dissociate into $ {K^ + } $ ion and $ C{l^ - } $ ion. Hence the value of $ i $ will be $ 2 $ .
Now, the depression in freezing point will be given as;
$ \Delta {T_f} = i \times {K_f} \times m $
$ \Delta {T_f} = 2 \times 1.8 \times 0.56 = 2.006\,K $
Hence the depression in the freezing point will be $ 2K $ .
Note:
Solutions containing non-volatile solute tend to show some properties that are only dependent on the concentration of solute particles and not on the nature of the solute. These properties are called colligative properties.
Complete Step-by-step Solution
The colligative properties of the solution do not depend on the chemical nature of the solution. In terms of solution, these properties tell us how the properties of a solution are connected to the solute concentration in the given solution. There are $ 4 $ colligative properties that are exhibited by the solutions. These are:
-Elevation in Boiling Point
-Depression in Freezing Point
-Relative Lowering in the Vapour Pressure
-Osmotic Pressure
Now, let’s talk about Depression in Freezing Point. According to Raoult’s law when you mix a non-volatile solute in a solvent it lowers the vapour pressure and becomes equal to the vapour pressure of the solvent at a lower temperature. This difference between the freezing point of pure solvent and that of a solution is called the depression in freezing point. It is given by the expression;
$ \Delta {T_f} = i \times {K_f} \times m $
Where, $ i = $ Van’t hoff factor
$ {K_f} = $ cryoscopic constant and $ m = $ molality of the solution
Now, we are given $ m = 0.56 $ , $ {K_f} = 1.8 $ . We will calculate the value of $ i $ and will put all the three values in the formula to find the value of $ \Delta {T_f} $ . As $ KCl $ is a strong acid it will dissociate into $ {K^ + } $ ion and $ C{l^ - } $ ion. Hence the value of $ i $ will be $ 2 $ .
Now, the depression in freezing point will be given as;
$ \Delta {T_f} = i \times {K_f} \times m $
$ \Delta {T_f} = 2 \times 1.8 \times 0.56 = 2.006\,K $
Hence the depression in the freezing point will be $ 2K $ .
Note:
Solutions containing non-volatile solute tend to show some properties that are only dependent on the concentration of solute particles and not on the nature of the solute. These properties are called colligative properties.
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