The properties of the solution are distinct compared to pure solvents and solutes. But several characteristics of the solution rely on the characteristics of the solute particles. It can be explained with some examples, the solution comprised of the solute Hydrogen chloride is acid while the solution of Ammonia is a base. And the solution of Sodium chloride is more packed while that of sucrose is thicker. But some characteristics of solutions do not rely on the character of solute particles existing instead it accounts only for the digit of solute particles present or the concentration of the solution. These characteristics rely only on the number of solute molecules is called colligative properties. Lowering of vapour pressure, depression in freezing point, etc. are some of the important colligative properties.
When the process of evaporation and condensation are arising at the same rate the pressure put forth by the vapours or gaseous particles present in the volatile solvent is its vapour pressure.
$$\mathrm{liquid\leftrightarrow gas}$$
The vapour pressure of a gaseous component is demonstrated in the below figure.
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But it has been observed that the addition of nonvolatile components in the solution is found to lessen the vapour pressure. The phenomenon of lowering vapour pressure by the addition of some nonvolatile solute is called relative lowering of vapour pressure.
The reason for this phenomenon is that in the method of vapourization solvent particles should be present on the exterior. But the presence of solute molecules decreases the availability of solute molecules on the exterior to escape as gas. Hence the equilibrium between liquid and gas is easily accomplished due to the lessening of vapour pressure. The below figure shows the reduction of vapour pressure by the addition of solutes.
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Raoult's law deals with the association between the lessening of vapour pressure and the concentration of solute particles.
The vapour pressure of the solution is directly proportional to the mole fraction of the solvent, $\mathrm{X_{A}}$. That is,
$$\mathrm{P\propto N_{A}/(N_{A}+N_{B})}$$
$$\mathrm{P=K[N_{A}/(N_{A}+N_{B})]}$$
$$\mathrm{P=KX_{A}}$$
Where, K = proportionality constant.
For a pure solvent, the value of K becomes equal to the $\mathrm{P^{0}_{A}}$, that is the vapour pressure of the pure solvent. That is,
$$\mathrm{P_{A}=X_{A}P^{0}_{A}}$$
$\mathrm{P_{A}}$ = Vapour pressure of component A
$\mathrm{P^{0}_{A}}$ = Vapour pressure of the pure solvent.
$\mathrm{X_{A}}$ = mole fraction of A
This law states that the partial pressure put forth by any particle of an ideal mixture is identical to the vapour pressure of the pure component multiplied by its mole fraction in the given solution.
Mathematical Representation
The mathematical representation of Raoult law is,
$$\mathrm{P_{A}=X_{A}P^{0}_{A}}$$
$\mathrm{P_{A}}$ = Vapour pressure of component A
$\mathrm{P^{0}_{A}}$ = Vapour pressure of the pure solvent.
$\mathrm{X_{A}}$ = mole fraction of A.
And that of relative lowering of vapour pressure is,
$$\mathrm{P_{A}^{0}-P_{A}/P^{0}_{A}=\Delta P_{A}/P^{0}_{A}}$$
At every range of temperature and concentrations, the ideal solution obeys Raoult's law. It is formed by the mixing of ideal solute(A) and ideal solvent(B). The intermolecular forces present in them are A-A and B-B. For ideal solutions, these intermolecular forces are almost equal. So they will over Raoult's law. Therefore, the vapour pressure is,
$$\mathrm{P_{A}=X_{A}P^{0}_{A}}$$
$$\mathrm{P_{B}=X_{B}P^{0}_{B}}$$
A mixture of Benzene and Toluene are examples of ideal solutions.
But some solutions do not obey this law and are nonideal solutions. They will not obey Raoult's law at all temperatures and concentrations. They will show positive and negative variations from Raoult's law. The equation for that solution that shows positive deviation is,
$$\mathrm{P_{A}>X_{A}P^{0}_{A}\:and\:P_{B}>X_{B}P^{0}_{B}}$$
In this case, the vapour pressure is greater than anticipated by Raoult law. Eg. Ethanol and Water.
And those who show negative deviation. The equation changed to,
$$\mathrm{P_{A}<X_{A}P^{0}_{A}\:and\:P_{B}<X_{B}P^{0}_{B}}$$
In this case, the vapour pressure is less than anticipated by Raoult law. Eg. Acetic Acid and pyridine.
With the help of relative lowering of vapour pressure, molar mass can be calculated. The molar mass of the solute can be found by following the steps given below.
$$\mathrm{X_{B}=\frac{N_{B}}{N_{A}+N_{B}}=\frac{W_{B}/M_{B}}{W_{A}/M_{A}+W_{B}/M_{B}}}$$
And
$$\mathrm{\frac{\Delta P_{A}}{P^{o}_{A}}=X_{B}=\frac{W_{B}/M_{B}}{W_{A}/M_{A}+W_{B}/M_{B}}}$$
Hence in this way molar mass is calculated.
The measurement of lowering of vapour pressure can be done using Walker and Ostwald dynamic method. By using this method, the individual determination of solute and solvent vapour pressure has been excluded. It consists of two bulbs in which the first bulb contains the solution and the second set contains pure solvents. And then they weighted accurately and connected. Then it is filled with a dehydrating agent such as Concentrated Sulfuric acid, Calcium chloride, etc. Then it is kept on a thermostat.
Then dry air is bubbled through it and then the bubbles become saturated with air. The blown air then takes vapours from the first bulb that is from the solution and then more from the second bulb that is the pure solvent. And is proportional to the disparity in vapour pressure of solvent and solution,$\mathrm{P^{o}-P}$.
$$\mathrm{The\:mass\:in\:the\:solution\:bulb\propto P}$$
$$\mathrm{The\:mass\:in\:the\:solvent\:of\:the\:bulb\propto P^{0}-P}$$
Then,
$\mathrm{P^{0}-P/P = loss of mass in solvent bulb/loss of mass in solution bulb.}$
Hence the relative lowering of vapour pressure can be calculated.
The properties that rely only on the number of solute particles and concentration instead of the identity of solute molecules are colligative. These properties are very important since they can be used for the determination of molar masses of solute particles. These are well observed when the solution is dilute and the solute is nonvolatile.
Four colligative properties are,
Reduction in freezing point.
Elevation in boiling point.
Osmotic pressure.
Lessening of vapour pressure.
The property that completely neglect neglects solute particles present in a solution is the colligative property. There are mainly four types. The reduction of vapour pressure is the result of the addition of solute particles to a particular solvent. Since the rate of vaporization is reduced by the addition of a nonvolatile solute. The law that governs the theory of relative lowering of vapour pressure is provided by Raoult's law. The main application of these colligative properties is molar mass determination. There are different methods and instruments for calculating the relative lowering of vapour pressure. Of which the important one is Walker and Ostwald method.
Q1. What is the connection between vapour pressure and the temperature of liquids?
Ans. They are directly related. When the temperature of liquid increases the vapour pressure will also increase. This is due to more molecules getting escaped from the liquid in the gaseous form. So the vapour pressure will increase.
Q2. What is the clapeyron Clausius equation?
Ans. The equation that shows the pressure-temperature relation is the Clausius clapeyron equation. The equation is,
$$\mathrm{ln(P_{1}/P_{2})=\Delta H_{vap}/R(1/T_{2}-1/T_{1})}$$
Where $\mathrm{P_{1}}$ and $\mathrm{P_{2}}$ are pressures at two temperatures $\mathrm{T_{1}}$ and $\mathrm{T_{2}}$.
Q3. What is the relation between density and rate of evaporation?
Ans. The rate of evaporation and density are inversely related. Since evaporation or escape of molecules in the form of gas has been reduced with higher density. Density is increased by the addition of more solutes.
Q4. Who discovered colligative properties?
Ans. Wilhelm Ostwald discovered the colligative properties of solute particles present in the solution.
Q5. Does lowering vapour pressure causes osmosis?
Ans. The experimental observations obtained by the occurrence of water in a solution placed in a sealed container with pure water as the source is due to lower vapour pressure of solution than pure water. And this is equivalent to osmosis where the air is the semipermeable membrane.
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