How to find Kf in Chemistry? - Molal freezing point depression constant
The symbol Kf represents the molal freezing point depression constant in Chemistry. It is formally known as the Cyroscopic constant.
Cryoscopic is derived from two Greek words, i.e., kyros, meaning frost and skopein, which denotes observation. Thus, Kf is related to observing the behavior of different chemical solutions under freezing conditions.
Depression in freezing point is a colligative property of matter. Thus, Kf helps in studying a colligative property, i.e., the effect of adding a particular amount of solute to a solvent on the freezing point of the solvent.
It refers to the decrease in the freezing point of a pure solvent when it is converted into a 1 molal solution by adding a non-volatile solute to it.
Molality refers to the amount of solute dissolved per kg of the solvent. A 1 molal solution of substance X in water denotes 1 mole of X dissolved completely in 1000 grams of water.
In Kf, K represents the equilibrium constant, while f denotes freezing point depression.
Kf is called a constant as different solvents have different Kf values. However, the Kf of a particular solvent stays fixed throughout. For instance, the Kf for water is1.86°C/m.
What are the units of Kf?
The units for Kf are Degrees Celsius (°C) per molal (m), i.e., °C/m.
As molality is calculated in mol/kg thus, the units for Kf can also be expressed as °C.kg.mol-1.
Conversely, the SI units for Kf are K.kg.mol-1.
Why is Kf important?
Kf denotes that there is a directly proportional relationship between depression in the freezing point and molality of the solution.
∆Tfα m
The greater the amount of solute added, the lower the freezing point of the solution.
Replacing proportionality with equality gives us:
∆Tf= Kf m
How to find Kf?
The molal freezing depression constant (Kf) can be determined using the formula ∆Tf = iKfm, where,
∆Tf = depression in freezing point where T stands for temperature (Unit: °C).
∆Tf = freezing point of the pure solvent (Tf solvent) – freezing point of the solution (Tf solution)
i = Van’t Hoff factor or ionization constant (No units).
It represents the number of particles or ions that the solute breaks down into when dissolved in a solvent.
Kf = Cyroscopic constant, aka molal freezing point depression constant of the solvent. (Units: °kg/mol or °C/m).
m = molality of the solution (Units: mol/kg or m).
The formula ∆Tf = iKfm can be rearranged to make Kf the subject of the formula:
Thus, if the values of ∆Tf, i and m are given, then we can easily find Kf using the equation Kf = ∆Tf/i.m.
If i = m = 1, then Kf = ∆Tf.
Experimentally, taking a non-dissociating, non-volatile solute and a specific solvent, chemists can determine the freezing point depression for different molality solutions.
Plotting a graph of ∆Tf versus molality, the value of Kf for the solvent can be inferred by taking the gradient of the graph (as shown below).
However, for now, let us apply the theoretical approach to finding Kf in the examples given below.
Examples of finding Kf using ∆Tf = iKfm
Example # 1: A 1 molal solution of NaCl in water undergoes a depression in freezing point (∆Tf) equal to 3.72°C. Determine the cryoscopic constant (Kf) of water using Kf = ∆Tf/i.m.
Solution:
As per the question statement:
Depression in freezing point (∆Tf) = 3.72°C
Molality (m) = 1 mol/kg
Ionization constant (i) = 2
1 NaCl molecule dissociates to produce 2 ions, i.e., Na+ and Cl–.
Cyroscopic constant = Kf =?
Result: The molal freezing point depression constant (cryoscopic constant) of water is 1.86°C/m.
Example # 2: A 0.625 molal solution of a non-dissociating compound is prepared in liquid benzene, which leads to a depression in freezing point (∆Tf) = 3.2°C. Use this information to calculate the molal freezing point depression constant (Kf) of benzene.
Solution:
As per the question statement:
∆Tf = 3.2°C
i = 1 (as the solute is a non-dissociating compound so no ions are produced in the solution)
m = 0.625 mol/kg
Kf = ?
Result: The molal freezing point depression constant (cryoscopic constant) of liquid benzene is 5.12°C/m.
Example # 3: 3.5 grams of calcium chloride (CaCl2) is completely dissolved in 100 grams of acetic acid. Find the value of Kf of acetic acid, given that the depression in the freezing point of the solution is 0.37°C.
Solution:
As per the question statement:
∆Tf = 0.37°C
1 CaCl2 molecule, upon complete ionization, releases 1 Ca+ and 2 Cl– ions.
i = 1 + 2 = 3
⇒ 3.5 grams of CaCl2 means:
∴ Moles = mass/molar mass
∴ Moles of CaCl2 = 3.5/110.98 = 0.0315
Molality is defined as the number of moles of solute dissolved per kg of solvent.
As per the above calculation, 0.0315 moles of CaCl2 are dissolved per 100 g of acetic acid.
Moles of CaCl2 in 1 kg i.e., 1000 g acetic acid = (0.0315/100)×1000 = 0.315 mol/kg
Result: The molal freezing point depression constant (cryoscopic constant) of acetic acid is 0.39°C/m.
Example # 4: Assuming 100% ionization, 0.6 molal solution of aluminum chloride (AlCl3) in cyclohexane leads to a depression in freezing point (∆Tf) = 322 K. Find the value of Kf for cyclohexane in °C.kg/mol.
Solution:
As per the question statement:
m = 0.6 mol/kg
∆Tf = 322 K
As Kf is asked for in °C.kg/mol., therefore to bring consistency in units, we need to convert ∆Tf from K to °C as shown below:
1 °C = 1K – 273
∆Tf = 322 – 273 = 49°C
1 AlCl3 molecule, upon 100% ionization, releases 1 Al3+ and 3 Cl– ions.
i = 1 + 3 = 4
Result: The molal freezing point depression constant (cryoscopic constant) of cyclohexane is 20.4°C.kg/mol.
FAQ
In chemistry, what is Kf?
Kf stands for molal freezing point depression constant or cryoscopic constant in Chemistry.
It represents the depression in the freezing point of a solvent if it is converted into a 1 molal solution by adding a non-volatile, non-dissociating solute into it.
∆Tf = i.Kf m
Kf = ∆Tf (if i = 1 and m = 1)
How to find Kf using the freezing point depression formula?
The freezing point depression formula is ∆Tf = i. Kf m.
The formula can be rearranged as Kf = ∆Tf/i.m where,
∆Tf = Depression in freezing point
i = Van’t Hoff factor or ionization constant
Kf = Molal freezing point depression constant or Cyroscopic constant
m = molality of the solution
Which of the following options is correct?
Kf for a solvent is:
A) Azeotropes
B) Ebullioscopic constant
C) Molal elevation constant
D) Cyroscopic constant
Option D is the correct answer. Cyroscopic constant is another name for the molal freezing point depression constant (Kf).
Why is depression in freezing point (∆Tf) a colligative property?
Colligative properties refer to the physical changes that occur as a result of adding a solute into a solvent. These properties only depend on the concentration of the solute, regardless of its chemical entity.
Depression in freezing point (∆Tf) is a colligative property as it directly depends on the amount of solute added, i.e., the molality of the solution.
∆Tfα m
∆Tf = i Kf. m
Where Kf = Molal freezing point depression constant.
Ammara Waheed is a highly qualified and experienced chemist, whose passion for Chemistry is evident in her writing. With a Bachelor of Science (Hons.) and Master of Philosophy (M. Phil) in Physical and Analytical Chemistry from Government College University (GCU) Lahore, Pakistan, with a hands-on laboratory experience in the Pakistan Council of Scientific and Industrial Research (PCSIR), Ammara has a solid educational foundation in her field. She comes from a distinguished research background and she documents her research endeavors for reputable journals such as Wiley and Elsevier. Her deep knowledge and expertise in the field of Chemistry make her a trusted and reliable authority in her profession. Let's connect - https://www.researchgate.net/profile/Ammara-Waheed
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