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Substances are a mixture of two or more pure substances. Molecular weight determination using solution colligative properties is one of the easiest methods to find molecular weight. Properties that are dependent on concentration of the solute ions of solute molecules are called colligative properties. In other words, we can say that the colligative property depends on the number of solutes and independent solute particles.
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Key terms: Vapour pressure, Osmotic pressure, Collision Theory, Boiling point, Freezing point, Mixture, Substance, Solution
Lowering in vapour pressure
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Raoult established that lowering in vapour pressure depends on the concentration of solute particles. If the vapour pressure of a solvent in a solution is less than that of the pure solvent & it is independent of their identity. The solute may or may not be volatile. The solutions contain two components, namely, the solutions of (i) liquids in liquids and (ii) solids in liquids.
Let us consider,
\(p_{1} = x_{1}p_{1}^{0}\)
\(\triangle p_{1} = p_{1}^{0} - p_{1} = p_{1}^{0} - p_{1}^{0}x_{1}\)
\(= p_{1}^{0}(1-x_{1})\)
\(\triangle p_{1} = x_{2} p_{1}^{0}\)
\(\frac{\triangle p_{1}}{ p_{1}^{0}}=\frac { p_{1}^{0} - p_{1}}{ p_{1}^{0}} = x_{2}\)
\(\frac { p_{1}^{0} - p_{1}}{ p_{1}^{0}} = \frac{n_{2}}{n_{1}+n_{2}} (\text {since }x_2 = \frac{n_{2}}{n_1+n_2})\)
\(\frac { p_{1}^{0} - p_{1}}{ p_{1}^{0}} = \frac{n_{2}}{n_{1}} \)
or we can also write it as
\(\frac { p_{1}^{0} - p_{1}}{ p_{1}^{0}} = \frac{w_{2} \times M_{1}}{M_{2} \times w_{1}} \)
The video below explains this:
Colligative Properties and Determination of Molar Mass Detailed Video Explanation:
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Osmotic pressure
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The pressure which stops the flow of solvent is called osmotic pressure. The osmotic pressure of a solution is the excess pressure that must be applied to a solution to prevent osmosis, i.e., to stop the passage of solvent molecules through a semipermeable membrane into the solution osmotic pressure is proportional to molarity, C of solution at a given temperature T. Π=CRT ( Π=osmotic pressure)
Π = (n2 /V) R T
Π V = w2 R T/ M2 OR
M2 = w2 RT/ Π V (knowing the quantities w2, T, Π and V. We can calculate the molar mass of the solute)
Elevation in boiling point
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An increase in boiling point Kb is called Boiling Point Elevation Constant or Molal Elevation Constant (Ebullioscopy Constant).
Elevation in boiling point
Elevation in boiling point is expressed in
ΔTb = Tb - TbO
\(\triangle Tb=Kbm\)
m= 1000 x w2/w1 x M2 m = 1000 x w1w2 x M2\(\triangle Tb = \frac{Kb \times 1000 \times w_2}{w_1 \times M_2}\)
\(\triangle Tb = Kb \times 1000 \times w_2w_1 \times M_2\)
Hence, the molecular weight will be
M2 = Kb x 1000 x w2/w1 x M2
Depression in freezing point
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The lowering of the vapour pressure of a solution causes a lowering of the freezing point. Raoult's law states that when non-volatile solid is added to solvent its vapour pressure decreased and it would become equal to that of solid solvent at lower temperature thus freezing point decreases.
Depression of freezing point (\(\triangle\)Tf) for dilute solution (ideal solution) is directly proportional to molality,
\(\triangle T_{t} \propto m\)
\(\triangle T_{t} = K_{f} m\)
\(\triangle T_{t} = \frac{K_{f} m \times 1000 \times w_2}{w_1 \times M_2}\)
\(\triangle T_{t} = (K_{f} \times 1000 \times w_2)(w_1 \times M_2)\)
Therefore equation becomes
\(M_{t} = \frac{K_{f} \times 1000 \times w_2}{w_1 \times \triangle T_f}\)
\(M_{2} = (K_{f} \times 1000 \times w_2)(w_1 \times \triangle T_f)\)
The proportionality constant k1 which depends on the nature of the solvent is known as the freezing point depression constant or molal depression constant or cryptic constant.
Depression in freezing point
From these formulas mentioned, we can find molecular weight with help of colligative properties.
Things to remember
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- If boiling is carried out in an open atmosphere, then external pressure is atmospheric pressure.
- The solution which obeys Raoult’s law over the entire range of concentration is called an ideal solution.
- If a solution does not obey Raoult’s law, the solution is non–ideal.
- The osmotic pressure is not the pressure produced by the solution.
- It exists only when the solution is separated from the solvent by the semipermeable membrane.
- The resulting osmosis produces an excess pressure (osmotic pressure) in the solution.
Also Read:
Sample Questions
Ques. Derive Expression of Raoult’s law for a solution of non-volatile solute or Derive Expression for relative lowering of the vapour pressure of a solution containing non-volatile solute? (5 marks)
Ans. Consider a solution of two components A1 and A2 with the mole fraction 12x and x respectively.
In this solution the component A2 (i.e. solute) is non-volatile, hence it does not evaporate and does not contribute to the total vapour pressure of the solution.
The vapour pressure of pure component A1 is p10 and that of component A2 is p10= 0
\(p_{T} = p^0_2 +(p^0_1 - p^0_2) x_1\)
\(\text{as }p^0_2 = \)
\(p_{T} = p^0_1 x_1\)
Thus, the vapour pressure of a solution is non-volatile solute is the product of vapour pressure of pure solvent p10 and mole fraction of the solvent, 1x which is Raoult’s law
For the component solution, x1 + x2 = 1 x1 = 1 – x2 where x2 is the mole fraction of non-volatile solute, not equal to zero. Hence x1<1.
Hence Product p10 x1 is always less than p10. Therefore, the vapour pressure of the solution, 0 To 1p p < which proves that lowering of the vapour pressure of a solution containing non-volatile solute
Lowering of vapour pressure is the product of vapour pressure of pure solvent and mole fraction of non-volatile solute dissolved in the volatile solvent to form a solution.
The lowering of vapour pressure depends on the nature of pure solvent and concentration of solute in mole fraction. The relative lowering of vapour pressure is given by,
\(\frac{\triangle P}{P_0^1} = \frac { P_{1}^{0} - P_{1}}{ P_{1}^{0}} = \frac{P_1^0x_{2}}{P_{1}^{0}} = x_{2}\)
Hence, relative lowering of vapour pressure=x (Mole fraction of non-volatile solute)
The above expression proves that the lowering of vapour pressure is a colligative property because it depends on the concentration of non-volatile solute.
Ques. How colligative properties are used in daily life? (2014, 2 marks)
Ans. Colligative properties used in daily life are given below:
- In an automobile used as anti-freezers with a low freezing point which helps operate automobile engines.
- Osmosis process is used in biological systems because of membranes semi-permeability nature.
- Osmosis is also used to remove waste products from the body.
Ques. Calculate the mole fraction of ethylene glycol C2H6O2 in a solution containing 20% of C2H6O2 by mass. (3 marks)
Ans. Assume that we have 100 g of solution (one can start with any amount of solution because the results obtained will be the same). The solution will contain 20 g of ethylene glycol and 80 g of water.
Molar mass of C2H6O2 = 12 × 2 + 1 × 6 + 16 × 2 = 62 g mol–1 .
Moles of C2H6O2 = 20g /62g mol = 0.322 mol
Moles of water=80 g/18 g=4.444 mol
xGlycol = moles of C2H6O2 / moles of C2H6O2 + moles of water
= 0.322/0.322 + 4.444 = 0.068
xWater = 4.444/0.322 + 4.444 = 0.932
mole fraction of water calculated by 1-0.628 = 0.932
Ques. Write the example of Henry’s law? (2 marks)
Ans. The following are the examples of Henry’s law:
- When a carbonated soft drink beverage bottle is seated with a cap, it is pressurized by a mixture of air and CO2.
- Due to the high partial pressure of CO2, the amount of CO2 in the dissolved state is high in soft drinks.
- When the cap is removed external pressure decreases, the solubility of CO2 decreases, and excess of CO2 and air in the bottle escapes out.
Ques. How Determine of Molar mass of non–volatile solute and the relative lowering of vapour pressure? (5 marks)
Ans. Let W2 g of solute of molar mass M2 be dissolved in W1 g of the solvent of molar mass M1.
Hence the number of moles of solvent, n1, and a number of moles of solute n2, in solution is given as,
the mole fraction of solute is given by
combining both equations
For dilute solutions, n1 n2 >>. Hence n2 may be neglected in comparison with n1 in equations that take the form.
Knowing the masses of non-volatile solute and the solvent in dilute solutions and by determining experimentally the vapour pressure of pure solvent and the solution It is possible to determine the molar mass of a non–volatile solute.
Ques. What are homogenous and heterogenous solutions? ( 2018, 4 marks)
Ans. The solution whose composition is uniform throughout the body of the solution is called a homogeneous solution.
- Formation/Preparation: The homogeneous solution is formed due to the force of attraction between the molecules or particles of solute and solvent.
- Heterogeneous solution: It is defined as the mixture of two or more phases.
- Solvation: It is defined as the process of interaction of solvent molecules with solute particles to form aggregates. When water is used as a solvent, it is called hydration or aquation.
Ques. What are colligative properties? (March 2013 4 marks)
Ans. The properties of solutions that depend only on the number of solute particles in solution and not on the nature of the solute particle are called colligative properties.
- Colligative properties are used to determine molar masses of non-electrolyte solutes.
- The relations derived by measuring colligative properties hold good for dilute solutions, with concentrations less than or equal to 0.2M.
There are Four Colligative Properties:
- Lowering of the vapour pressure of the solvent in the solution
- Elevation of the boiling point of the solvent in the solution
- Depression of freezing point of the solvent in the solution
- Osmotic pressure
Ques. Show graphical representation of elevation of the boiling point of the solvent in solution or Show Variation of the vapour pressure of pure solvent and solution with temperature. (5 marks)
Ans.
- The vapour pressure-temperature curve of the solution is always below the vapour pressure-temperature curve of the pure solvent.
- The boiling point of pure solvent is 0 T and that of the solution is T.
- The elevation of boiling point ΔTb is represented by the distance AB. (ΔTb=− T -T0)
- The lowering of vapour pressure (P10 -P) at the temperature T0 is equivalent to AC.
- The elevation of boiling point is proportional to the lowering of vapour pressure ΔT ∝ Δp
Ques. What are the types of solutions on the basis of osmotic pressure? (2012, 6 marks)
Ans. There are three types of solution on the basis of osmotic pressure that is given below:
- Isotonic solution -
- Two or more solutions exerting the same osmotic pressure are called isotonic solutions.
- For example, ( ) 1 0.05M 3.0 gL− urea solution and ( ) 1 0.05M 17.19 gL− sucrose solution are isotonic because their osmotic pressures are the same.
- If these solutions are separated by a semipermeable membrane, there is no flow of solvent in either direction.
- Hypertonic solution -
- A solution having osmotic pressure higher than that of another solution is said to be hypertonic with that solution.
- E.g., 0.1 M urea solution exerts higher osmotic pressure than 0.05 M sucrose solution. Hence, 0.1M urea solution is hypertonic to 0.05 M sucrose solution.
- If these solutions are separated by a semipermeable membrane, the solvent flows from sucrose to urea as sucrose is having low concentration.
- Hypotonic solution -
- A solution having osmotic pressure lower than that of another is said to be a hypotonic solution with that solution.
- For example, 0.05 M sucrose solution has osmotic pressure lower than that of 0.1 M urea solution. Therefore 0.05 M sucrose solution is hypotonic with 0.1 M urea solution.
Ques. The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. Kb for benzene is 2.53 K kg mol–1 (2 marks)
Ans. The elevation (\(\triangle T_b\)) in the boiling point,
= 354.11 K – 353.23 K
=0.88 K
Substituting these values in an expression, we get
M2 = 2.53 K kg mol-1 × 1.8 g × 1000 g kg-1/ 0.88 K × 90 g
= 58 g mol-1
Therefore, the molar mass of the solute, M2 is 58 g mol-1.
Ques. 200 cm3 of an aqueous solution of a protein contains 1.26 g of the protein. The osmotic pressure of such a solution at 300 K is found to be 2.57 × 10-3 bar. Calculate the molar mass of the protein. (2 marks)
Ans. The various quantities known to us are as follows Π=2.57 × 10–3 bar,
V = 200 cm3
= 0.200 litre
T = 300 K R
=0.083 L bar mol-1 K-1
Substituting these values in equation, we get
M2 = 1.26 g × 0.083 L bar K-1 mol-1 × 300 K/ 2.57×10-3 bar × 0.200 L
= 61,022 g mol-1
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