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Depression in Freezing Point - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:35 AM | #JEE Main

Quick Facts

  • Depression in Freezing Point is considered one of the most asked concept.

  • 15 Questions around this concept.

Solve by difficulty

K for water is 1.86K kg mol -1   If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol (C2H6O2) must you add to get the freezing point of the solution lowered to –2.8o C?

Concepts Covered - 1

Depression in Freezing Point

Depression in Freezing Point:

Freezing Point: It is the temperature at which the liquid and the solid form of the same substance are in equilibrium and have the same vapour pressure. A solution freezes when its vapour pressure is equal to the vapor pressure of pure solid solvent. Due to lower vapour pressure of the solution, solid form of a solution separates out a lower temperature.

On adding a non-volatile solute to the solvent, the vapor pressure of the solution is lesser than the solvent and the vapor pressure of the solution becomes equal to the vapor pressure of solid solvent at a lower temperature when compared to the pure solvent and hence the freezing point of decreases.

 

Suppose \mathrm{T_{f}^{0}\ and \ T_{f}} are the freezing point of pure solvent and solution respectively. Decrease in freezing point \Delta T_{f} is given as:

\mathrm{\Delta T_{f}= T_{f}^{0}-T_f}

  • This is also termed as cryoscopy and depression of freezing point (\Delta T_f)
  • For a dilute solution, \Delta T_f is directly proportional to the molality (m) of the solution.

            Hence \mathrm{\Delta T_f\propto m}
            \Delta T_f=K_f m
            If molality of the solution is one, then
              \Delta T_f=K_f
            \Delta T_f and M can be found out by using these relations.

            \\\mathrm{\Delta T_{f}=K_{f}\times \: \frac{w}{M} \times \frac{1000}{W}}\\\\\mathrm{M=\frac{K_{f} \times w \times 1000}{\Delta T_{f} \times W}}

 

Here w = weight of solute
W = weight of solvent
Kf = molal depression constant or cryoscopic constant.
M = molar mass of solute

M1 = molar mass of solvent

The value of Kor Cryoscopic constant is a property of the solvent only and does not depend on the type of solute. The value of Kcan be calculated as: 
\\\mathrm{K}_{\mathrm{f}}=\frac{\mathrm{M_{1}RT}^{2}}{1000 \mathrm{L}_{\mathrm{f}} \text { or } \Delta \mathrm{H}_{\text {fusion }}}


Here, \mathrm{L_f \: or\: \Delta H_f \textrm{ = latent heat of fusion}}

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Depression in Freezing Point

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