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Freezing point of a substance can be defined as the temperature at which the vapour pressure of the substance in its liquid phase is equal to its vapour pressure in the solid phase. According to Raoult’s Law, the vapour pressure of a pure solvent decreases when a non-volatile solute is dissolved in it. From these two consecutive statements, we can conclude that the freezing point of the solvent decreases.
The Depression of Freezing Point can be defined as the decrease in the freezing point of a solvent when a solute is added to it.
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Keyterms: Freezing Point Depression, liquid, solid phase, vapour pressure, substance, solvent, Raoult’s Law, freezing point, non-volatile solute
Calculation of the Depression of Freezing Point
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Consider a pure solvent whose freezing point is denoted by Tf°. Now, add the non-volatile solute into the solvent. Let the freezing point of the solution be Tf.
According to the Depression of Freezing Point,
Tf < Tf°
This difference in the freezing point of the pure solvent and the solution can be denoted by ΔTf.
Therefore, we get
ΔTf = Tf° - Tf
where
ΔTf is the Depression of Freezing Point
Tfo is the freezing point of the pure solvent
Tf is the freezing point of the solution
Calculation of the Depression of Freezing Point for Dilute Solution
The Depression of Freezing Point for a dilute solution is directly proportional to the molality(m) of the solution.
ΔTf ∝ m
Therefore,
ΔTf = Kf . m . i
ΔTf is the Depression of Freezing Point
Kf is the Freezing Point Depression Constant or Cryoscopic Constant
m is the molality of the solution
i is the Van’t Hoff Factor
The Van't Hoff Factor is the number of ion particles per formula unit of solute. For example, the Van ‘t Hoff Factor of NH3, NaCl, and Ca(NO3)2 is 1, 2, and 3 respectively.
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| Related Concepts | ||
|---|---|---|
| Cis Trans Isomerism | Azeotropic Distillation | Atomic Mass of Elements |
| Mass Percent Formula | Melting and Boiling Point | Mole Fraction |
Freezing Point Depression Constant or Cryoscopic Constant
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Kf is defined as the depression in freezing point when 1 mole of solute is dissolved in 1000g of the solvent. It is the characteristic of a particular solvent. It can be calculated from the following relation:
Kf = R x M1 x Tf ^ 2 / 1000 x ΔfusH
Kf is the Cryoscopic Constant
R is the Gas Constant
M1 is the molar mass of the solvent
Tf is the freezing point of the pure solvent in Kelvin
ΔfusH is the Enthalpy of Fusion of the solvent
Calculating Molar Mass of the Solute from the Depression of Freezing Point
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Consider a solution in which w2 gram of solute with a molar mass of M2 is mixed in w1 gram of solvent. The depression in the freezing point of the solvent is ΔTf.
Now, the molality of the solute is given by,
m = [w2 / M2] / [w1 / 1000]
Substituting ‘m’, we get
ΔTf / Kf = [w2 / M2] / [w1 / 1000]
ΔTf = Kf x [w2 / M2] / [w1 / 1000]
ΔTf = Kf x w2 x 1000 / M2 x w1
Taking M2 the other side, we get
M2 = Kf x w2 x 1000 / ΔTf ? w1
M2 is the molar mass of the solute
Kf is the Cryoscopic Constant
w2 is the amount of solute added
ΔTf is the Depression of Freezing Point of the solvent
w1 is the amount of solvent
Uses of the Depression of Freezing Point
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There are many practical uses of the Depression of Freezing Point. These uses are listed below.
- The radiator fluid in automobiles is a mixture of water and ethylene glycol to prevent the freezing of radiators in winter. The addition of ethylene glycol lowers the freezing point of water.
- Salt is poured over roads (called Road Salting) in winters. This lowers the freezing point and prevents the accumulation of dangerous ice that can cause accidents.
- Many organisms have developed a coping mechanism to survive in extremely cold conditions. They produce a high concentration of certain substances that lowers the freezing point of water inside them, enabling them to survive in cold conditions.
Things to Remember
- The freezing point of the solution is always less than the freezing point of the pure solvent.
- The freezing point for the dilute solution is calculated by: ΔTf = Kf . m . i
- Molar mass of the solute is calculated by: m = [w2 / M2] / [w1 / 1000]
- The process of lowering the freezing point finds many applications in day-to-day life activities.
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Sample Questions
Q-1 An unknown compound measuring 5g reduces the freezing point of 50g benzene from 5.53oC to 4.90oC. What is the molar mass of the unknown compound? Kf = -5.12oC/m (2 marks)
Ans. w1 = 50g, w2 = 5g, Kf = -5.12oC/m, ΔTf = 4.90oC - 5.53oC = -0.63oC
We know that,
M2 = Kf x w2 x 1000 / ΔTf x w1
M2 = -5.12 x 5 x 1000 / -0.63 x 50
M2 = 812.70 g/mol
Q-2 Two elements A and B form compounds having formulas AB2 and AB4. 1 g of AB2 and AB4 lower the freezing point of 20 g of benzene (C6H6) by 2.3 K and 1.3 K respectively. Calculate atomic masses of A and B, if the molar depression constant for benzene is 5.1 K kg mol–1. (3 marks)
Ans. Using the formula,
M2 = Kf x w2 x 1000 / ΔTf x w1
We get,
M(AB2) = 5.1 x 1 x 1000 / 2.3 x 20 = 110.87 g/mol
M(AB4) = 5.1 x 1 x 1000 / 1.3 x 20 = 196.15 g/mol
Now, let the atomic masses of A and B be x and y.
AB2 = x + 2y = 110.87 … (1)
AB4 = x + 4y = 196.15 … (2)
Subtracting (2) from (1),
2y = 85.28
y = 85.28 / 2
y = 42.64
Therefore,
x = 110.87 - 85.28 = 25.69
Hence, the atomic mass of A = 25.69 g/mol and B = 42.64 g/mol.
Q-3. An aqueous solution of sodium chloride freezes below 273 K. Explain the lowering in freezing points of water with the help of a suitable diagram. (Comptt. Delhi 2013, 2 marks)
Ans. An aqueous solution of sodium chloride freezes below 273 K because vapour pressure g of the solution is g less than that of the pure solvent.
Q-4. What mass of NaCl (molar mass = 58.5 g mol-1) must be dissolved in 65 g of water to lower the freezing point by 7.5°C? The freezing point depression constant, Kf, for water is 1.86 K kg mol-1. Assume van’t Hoff factor for NaCl is 1.87. (All India 2010, 3 marks)
Ans. Given : M2 = 58.5 g mol-1 w1 = 65 g
ΔTf = 7.5 °C K, = 1.86 Kf kg mol-1 i = 1.87
Substituting these values in the formula
∴ Mass of NaCl to be dissolved, w2 = 8.199 g
Q-5. Some ethylene glycol, HOCH2CH2OH, is added to your car’s cooling system along with 5 kg of water. If the freezing point of a water-glycol solution is -15.0°C, what is the boiling point of the solution?
(Kb = 0.52 K kg mol-1 and Kf = 1.86 K kg mol-1 for water) (Comptt. Delhi 2014, 3 marks)
Ans. Given : ΔTf = 0 – (-15) = +15° C, w1 = 5 kg = 5000 g
Kf = 1.86 K kg mol-1 Kb = 0.52 K kg mol-1
Q-6. A 10% solution (by mass) of sucrose in water has freezing point of 269.15 K. Calculate the freezing point of 10% glucose in water, if freezing point of pure water is 273.15 K.
Given: (Molar mass of sucrose = 342 g mol-1) (Molar mass of glucose = 180 g mol-1) (Delhi 2017, 3 marks)
Ans.
Molality (m) = 
Given:
Molar mass of sucrose
= C12H22O11 = 12 × 12 + 22 + 11 × 16 = 342
10% solution (by mass) of sucrose in water means 10 g of sucrose is present in (100 – 10)
= 90 g of water
10% solution of sucrose means, w = 10 g
Mass of water, W = 90 g
∴ ΔTf for glucose = 12.33 × 0.6166 = 7.60 K (approx.)
∴ Freezing point of 10% glucose solution
= (273.15 – 7.60) K = 265.55 K
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