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Colligative Properties and Determination of Molar Mass
Relative Lowering of Vapour Pressure
Elevation of Boiling Point
Depression of Freezing Point
Osmosis and Osmotic Pressure
Applications of Osmosis
When we a non-volatile solute in a volatile solvent we observe that there is a decrease in vapour pressure of the solution. This decrease of vapour pressure can be quantitatively used to measure several properties of liquid solutions.These properties depend more on solute particles of the solution and are called colligative properties. The word “colligative” is derived from the Latin word “coligare” which means “to bind together”. The following properties come in the category of colligative properties.
Image 1: There is a relative lowering of vapour pressure because less surface area is available for the solvent molecules to escape out.
The vapour pressure of a solvent is lowered when we add a non-volatile solute in it; this is called Lowering of Vapour Pressure. The French chemist François-Marie Raoult’s observed that the concentration of solute particles is mainly responsible for the lowering of vapour pressure and also discovered a relation between vapour pressure of solution, vapour pressure of pure solvent and mole fraction of solute and solvent.
Image 2: The vapour pressure decreases as we add solute in pure solvent
If the vapour pressure of the solution is P_{1}, the vapour pressure of the pure solvent is P_{1}^{0} and mole fraction of solvent is x_{1}, then according to Raoult’s Law:
P_{1} = P_{1}^{0} x_{1}
The decrement in vapour pressure of solvent that is, Δ P_{1} will be
Decrement in Vapour Pressure = Vapour Pressure of Pure Solvent – Vapour Pressure of Solvent
Δ P_{1} = P_{1}^{0} – P_{1}
Substituting P_{1} = P_{1}^{0} x_{1} in the above relation we get,
Δ P_{1} = P_{1}^{0} – P_{1}^{0} x_{1}
Δ P_{1} = P_{1}^{0} (1 – x_{1})
Since the sum of mole fraction of solute ( x_{2}) and mole fraction of solvent ( x_{1}) is 1, we can write
1 – x_{1} = x_{2}
Δ P_{1 }= P_{1}^{0} x_{2}
It is obvious that decrease in vapour pressure depends on mole fraction of solute (x_{2}) as mentioned by Raoult’s.
The equation can be re-written as:
The term is called Relative Lowering in vapour pressure and equals to mole-fraction of the solute.
If n_{1} and n_{2} are the respective moles of solute and solvent, then we can re-write the equation as:
If the solution is quite diluted, we can neglect moles of solute n_{1 }in front of moles of solvent n_{2}. So, in case of dilute solution, the equation becomes
We know that moles of any substance can be calculated by dividing the given mass by its molecular mass. Then moles of solute (n_{2}) is equal to
where m_{2} and M_{2} are given mass and molecular mass of solute.
Similarly, a mole of solvent (n_{1}) equals
Substituting the value of n_{1} and n_{2} in the dilute solution vapour pressure relation, we get
So, we know the other quantities, we can easily determine the molar mass of solute M_{2} with the above relation.
Example: The vapour pressure of pure benzene is 0.850 bars at room temperature. On addition of 0.6 g of non-volute solute in 39 g of benzene solution, the vapour pressure of benzene reduces to 0.845 bars. What is the molar mass of solute?
We have
P_{1}^{0} = 0.850 bar, P_{1} = 0.845 bar, m_{1} = 39 g, M_{1} = 78 g and m_{2} = 0.6 g. Now the unknown quantity is M_{2} which is the molar mass of non-volatile solute.
Using the relative lowering of vapour pressure relation, we get:
The change in vapour pressure Δ P_{1} = P_{1}^{0 }– P_{1}, equals to 0.850 – 0.845 = 0.05 bar. Substituting the respective values in the relation we get
Calculating, we get the value of M_{2} as 20.68 g.
Image 2: There is a relative lowering of vapour pressure because less surface area is available for the solvent molecules to escape out.
We know that on addition of non-volatile solute in a solvent, the vapour pressure of the solvent decreases. However the boiling point of the solution is greater than that of the pure solvent, this is because vapour pressure is directly proportional to temperature. And in order to boil a solution, we need to raise the boiling point of the solution to a certain temperature. This raising of temperature is called elevation of boiling point and just like the relative lowering of vapour pressure it also depends on solute particles in the solution.
Mathematically, if T_{b}^{0} is the boiling point of pure solvent and T_{b} denotes boiling point of the solution, then elevation in boiling point (denoted by Δ T_{b}) is
Δ T_{b} = T_{b} – T_{b}^{0}
Experimentally it has been found that elevation in boiling point in dilute solutions is directly proportional to molality ‘m’ of solute present in a solution.
Δ T_{b} ∝ m
Δ T_{b} = K_{b }m
The term molality ‘m’ denotes the number of moles of solute present in 1000 g or 1 kg of solvent. In the relation K_{b} is the called molal elevation constant or ebullioscopic constant. The standard unit of molal elevation constant K_{b} is K kg mol^{-1}.
Let m_{1} and m_{2} be the given masses of solvent and solute respectively. And molar masses of solute be M_{2} and that of solvent is M_{1}, then molality can be evaluated from the relation:
Putting the value of molality in the boiling elevation relation, we get
Hence, if know the remaining quantities we can easily determine the molar mass of solute with the boiling elevation relation.
Example: The boiling point of a pure liquid is 353.23 K. If we add 2.70 g of a non-volatile solute in 90 g of liquid, the boiling point of the solution rises to 354.11 K. What will be the molar mass of non-volatile solute? Take the value of K_{b} of liquid to be 2.53 K kg mol^{-1}.
So we have
T_{b}^{0} = 353.23 K, T_{b} = 354.11 K, m_{2} = 2.70 g, K_{b} = 2.53 K kg mol^{-1} and m_{1} = 90 g.
The elevation in boiling point Δ T_{b} = T_{b} – T_{b}^{0} = 354.11K – 353.23K =0.88
Using the above relation we get:
Calculating by we get the value of M_{2} to be 86.25 g, hence molar of the mass of solute is 86.25 g.
When the vapour pressure of a solution is reduced, the freezing of the solution decreases. The freezing point of a solution is defined as the temperature at which vapour pressure of its substance become equal in liquid and vapour phase. If the vapour pressure of the solution equals to vapour pressure of the pure solvent, then the solution will be frozen. According to Raoult’s Law when we add a non-volatile solute in a solvent we find that the freezing point of the solution is slightly less than that of the pure solvent. This is called depression of freezing point.
The freezing point depression is denoted by Δ T_{f} and equals to
Δ T_{f} = T_{f}^{0} - T_{f}
The freezing point depression for dilute solutions is directly proportional to molality of the solute, just like the boiling elevation point. That is
Δ T_{f} ∝ m
Δ T_{f} = K_{f} m
The proportionality constant K_{f} is called Molal Depression Constant and is also known as Cryoscopic Constant, which depends upon the nature of the solvent.
We know that molality ‘m’ equals to
where m_{2} is a mass of solute, m_{1} is a mass of solvent and M_{2} is the molar mass of added non-volatile solute.
Putting the value of molality in above equation we get depression in freezing point as
Hence, we can use depression of freezing point to evaluate molar mass of solute.
Image 3: A brain map to remember how to calculate the molar mass of solute when depression in freezing point is given.
The table below depicts boiling and freezing points of important solvents with their molal constants.
Solvent
Boiling Point (K)
K_{b} ( K kg mol^{-1})
Freezing Point (K)
K_{f} ( K kg ol^{-1})
Water
373.15
0.52
273
1.86
Ethanol
351.5
1.20
155.7
1.99
Cyclohexane
353.74
2.79
279.55
20
Benzene
353.3
2.53
278.6
5.12
Chloroform
334.4
3.63
209.6
4.79
Acetic Acid
391.1
2.93
290
3.90
Diethyl Ether
307.8
2.02
156.9
1.79
Image 4: Osmosis process
Osmosis is defined as the flow of liquid through a semi-permeable membrane which allows only solvent molecules to pass through it. The term membrane can be well-understood from real life examples like blood cells get destroyed when placed in a salt-water solution, raw mangoes get shrined when we prepare pickles with brine solution etc. A membrane is a continuous sheet or film with a pore network embedded in such way that it allows only desired substances to flow through it. It can be natural like a pig’s bladder or it can be man-made as cellophane is.
The membrane which allows only small solvent molecules to pass through it and resist the flow of bigger solute molecules is called Semipermeable Membrane or SPM.
Osmotic Pressure is the defined as the extra pressure one can add to stop the solvent molecules to flow. In other words, the pressure exerted on a solution to prevent osmosis with the help of semi-permeable membrane is called Osmotic Pressure. Osmotic pressure is a colligative property and depends on solute particles of the solution. Experimentally for dilute solutions, osmotic pressure is
π = CRT
where π is osmotic pressure, R is gas constant, T is constant and C denotes concentration or molarity of the solution.
Molarity (C) is defined as moles of solute divided by the volume of solution in liters. Mathematically
C = n_{2} / V
where n_{2} is moles of solute and V is the volume of the solution. Putting the value of c in osmotic pressure we get
π = (n_{2} / V) R T
Since moles n_{2} = m_{2} /M_{2}, where m_{2} is given mass and M_{2} is molar mass, we can derive a new relation between osmotic pressure and molar mass of solute M_{2} as
Thus, if we know the remaining quantities we can easily determine the molecular masses of solute. In fact, this method is used to determine molecular masses of several biomolecules, proteins, and polymers. The advantage of using osmotic pressure is that we use molarity of the solution instead of molality at standard room temperature.
We can also determine density of the solution with the relation
π = d g h
where π is osmotic pressure, d is density of the solution, g is acceleration due to gravity and h is height of the liquid column.
Image 5: Relation between osmotic pressure and density of the solution
Notes:
Two solutions which have same osmotic pressure are said to be isotonic solutions and when we separate these solutions by an SPM, osmosis never occurs
The solution with higher osmotic pressure as compared to other solution is called Hypertonic Solution
The solution with lower osmotic pressure as compared to other solution is called Hypotonic Solution
Osmosis process is of great utility in commercial and daily life. It has following applications:
Reverse Osmosis, (a process in which we reverse the direction of osmosis by applying larger pressure than osmotic pressure on the other side, which help in getting pure solvent out of the solution) is used to obtain fresh water from the sea water
Image 6: We add cellulose acetate to act as semipermeable membrane
Also, we use cellulose acetate membrane to obtain fresh water as it only allows pure water solvent molecules and is impermeable to impurities and ions present in contaminated water
Watch this Video for more reference
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