Notes On Colligative Properties: Depression Of Freezing Point - CBSE Class 12 Chemistry
The freezing point of a substance is defined as the temperature at which its solid phase is in dynamic equilibrium with its liquid phase. At the freezing point, the vapour pressure of the substance in its liquid phase is the same as the vapour pressure of the substance in its solid phase. When a non-volatile solute is added to a solvent, the freezing point of the solution gets lowered. According to Raoultâ€™s law, the vapour pressure of a solution containing a non-volatile solute is lower than that of the pure solvent. Thus freezing point of a solvent decreases when a non-volatile solute is added to it. The depression in freezing point depends upon the concentration of the solution. For dilute solutions, depression in the freezing point is directly proportional to molality (m). Thus, âˆ†Tf =Kf m Where Kf =freezing point depression constant (or) molal depression constant (or) cryoscopic constant. Molal depression constant Kf can be defined as the depression in freezing point when 1mole of solute dissolved in 1kg of solvent. The unit for Kf is kelvin kilogram /mole. As Kf depends upon the nature of the solvent, its value is different for different solvents. The values of Kf can be calculated from this expression Kf = (R x M1 x Tf2)/(1000 x Î”fusH)     R = Gas constant    M1= Mlar mass of the solvent    Tf  = Freezing point of the pure solvent Î”fusH = Enthalpy for the fusion of the solvent If w2 grams of a solute with molar mass M2 is dissolved in w1 grams of a solvent, then molality m of the solution is given by W2 multiplied by 1,000 divided by w1 multiplied by M2 Substituting this value of molality in the freezing point depression equation, we get depression in freezing point Molarity , m = (W2 x 1000)/(W1xM2)          Î”Tf = Kf m          Î”Tf = (Kf x W2 x 1000)/(W1xM2)               M2 = = (Kf x W2 x 1000)/(W1xÎ”Tf) Thus, the molar mass of a non-ionic solute can be calculated by using the depression in freezing point.   Important applications of depression in freezing point:  Running a car in sub-zero temperatures even when the radiator is full of water is possible due to the fact that a depression in the freezing point of water takes place when an appropriate amount of solute (ethylene glycol) called anti-freeze is dissolved in it. The addition of the solutes like Sodium chloride depresses the freezing point of water to such an extent that it cannot freeze at the prevailing temperature and hence the snow melts off easily.

#### Summary

The freezing point of a substance is defined as the temperature at which its solid phase is in dynamic equilibrium with its liquid phase. At the freezing point, the vapour pressure of the substance in its liquid phase is the same as the vapour pressure of the substance in its solid phase. When a non-volatile solute is added to a solvent, the freezing point of the solution gets lowered. According to Raoultâ€™s law, the vapour pressure of a solution containing a non-volatile solute is lower than that of the pure solvent. Thus freezing point of a solvent decreases when a non-volatile solute is added to it. The depression in freezing point depends upon the concentration of the solution. For dilute solutions, depression in the freezing point is directly proportional to molality (m). Thus, âˆ†Tf =Kf m Where Kf =freezing point depression constant (or) molal depression constant (or) cryoscopic constant. Molal depression constant Kf can be defined as the depression in freezing point when 1mole of solute dissolved in 1kg of solvent. The unit for Kf is kelvin kilogram /mole. As Kf depends upon the nature of the solvent, its value is different for different solvents. The values of Kf can be calculated from this expression Kf = (R x M1 x Tf2)/(1000 x Î”fusH)     R = Gas constant    M1= Mlar mass of the solvent    Tf  = Freezing point of the pure solvent Î”fusH = Enthalpy for the fusion of the solvent If w2 grams of a solute with molar mass M2 is dissolved in w1 grams of a solvent, then molality m of the solution is given by W2 multiplied by 1,000 divided by w1 multiplied by M2 Substituting this value of molality in the freezing point depression equation, we get depression in freezing point Molarity , m = (W2 x 1000)/(W1xM2)          Î”Tf = Kf m          Î”Tf = (Kf x W2 x 1000)/(W1xM2)               M2 = = (Kf x W2 x 1000)/(W1xÎ”Tf) Thus, the molar mass of a non-ionic solute can be calculated by using the depression in freezing point.   Important applications of depression in freezing point:  Running a car in sub-zero temperatures even when the radiator is full of water is possible due to the fact that a depression in the freezing point of water takes place when an appropriate amount of solute (ethylene glycol) called anti-freeze is dissolved in it. The addition of the solutes like Sodium chloride depresses the freezing point of water to such an extent that it cannot freeze at the prevailing temperature and hence the snow melts off easily.

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