شرح درس
Collective Properties of Solutions
CHEMISTRY - Secondary 1
In This part we will study
 Freezing point depression :
* The freezing point of solution is less than the freezing point of pure solvent which forms it, as the attraction forces between the solvent molecules and the solute molecules in the solution hinders the conversion of the solvent from liquid state to solid state (crystals), hence it is required to decrease the temperature of the solution to less than the freezing point of the pure solvent, so that the solute crystals separate from the solvent crystals.

*Freezing point depression is directly proportional to the number of dissolved solute particles in solution, where the freezing point of a solution decreases with increasing its molal concentration.

*When 1 mol of any nonelectrolyte such as glucose is dissolved in 1 kg of water the resulting
solution freezes at (-l .86°C).
,while the substance which is dissociated into ions in water (electrolyte), its freezing point is
calculated through the relation :

The freezing point of electrolytic solution = Number of moles of ions in the solution × (-l.86°C)

Example :
What is the freezing point of the solution, that contains 1 mol of calcium chloride CaC12 in 1000 g of water ?
Solution

Notes
 Decreasing in the freezing point of sodium chloride solution is double the decreasing in the freezing point of glucose solution which has the same molal concentration.

As the dissolving of 1 mol of glucose sugar in water forms 1 mol of its molecules in the solution, while dissolving of 1 mol of NaCl in water forms 2 mol of ions in the solution, which leads to double decreasing the freezing point.


 To prevent cars from slipping and decrease the number of accidents where salt dissolves in water forming a solution, which freezes at a temperature below the freezing point of pure water, So the amount of snow will decrease.

Comparison between the solution and the pure solvent "in terms of: colligative properties"

* The change in the boiling point of the solution depends on :
- The molal concentration of the solute (msolute) in the solution.
- The molal boiling point elevation constant (Ebullioscopic constant) (Kb).
* The change in the freezing point of the solution depends on :
- The molal concentration of the solute (msolute) in the solution.
- The molal freezing point depression constant (Cryoscopic constant) (Kf).
* The following table illustrates the values of Kb, Kr, the boiling points and the freezing points of different solvents.

* The elevation in the boiling point is estimated through the relation :
∆T = Kbmsolute

* The depression in the freezing point is estimated through the relation :
∆T = Kf msolute
Example (1) :
Calculate the elevation in the boiling point of carbon tetrachloride (its molal boiling point elevation constant is 5.03 °C.kg/mol) when 2 g of the substance X
(its molar mass = 500 g/mol) is dissolved in 15 g of it.
Solution

Example (2) :
Calculate the freezing point of the solution which is produced from dissolving 2 g of a given compound (its molar mass = 125 g/mol) in 35 g of camphor, knowing that the freezing point of camphor is 178.4 °C and its molal freezing point depression constant is 40 °C. kg/mol
Solution

Best wishes to you
Mr.Ahmed Elbasha

كيمياء الصف الاول الثانوي لغات - الفصل الدراسي الاول
شرح كيمياء لغات اولي ثانوي - مستر احمد الباشا
كيمياء لغات - الصف الاول الثانوي - كيمياء اولي ثانوي الترم الاول - كيمستري اولي ثانوي الترم الاول