Lowering of Vapour Pressure: Relative lowering of vapour pressure is a colligative property of solutions. Students can find this chapter in their CBSE class 12 science syllabus. Relative lowering of vapour pressure in solutions stands for the ratio of lowered vapour pressure to the vapour pressure of the pure solvent. Where P1 and P2 are the respective vapour pressures of the solvent and the solute. Since the solute is non-volatile, its contribution to the total vapour pressure will be zero, i.e. P2=0.
Let us discuss the relative lowering of vapour pressure in the colligative solutions property with more detail and emphasis. This article will contain the definition of a reduction in vapour pressure and the other colligative properties too.

Define Lowering of Vapour Pressure

The vapour pressure of a substance is defined as the pressure exerted by its vapour state on its liquid (or solid) state in a closed container at equilibrium conditions. The vapour results from the evaporation of the liquid or solid sample. At a given temperature, this vapour pressure is in equilibrium with its solid or liquid state.

Examples:

Substance	Vapour Pressure at \({\rm{25}}{\,^{\rm{o}}}{\rm{C}}\)
diethyl ether	\({\rm{0}}.{\rm{7}}\,{\rm{atm}}\)
bromine	\({\rm{0}}.{\rm{3}}\,{\rm{atm}}\)
ethyl alcohol	\({\rm{0}}.{\rm{08}}\,{\rm{atm}}\)
water	\({\rm{0}}.{\rm{03}}\,{\rm{atm}}\)

Characteristics of Vapour Pressure

1. Vapour pressure results when a solid or a liquid evaporates to its gaseous state in a closed container. In a closed container, the gaseous molecules cannot escape; hence they exert pressure on their solid or liquid state.
2. During evaporation, some gas molecules will eventually strike the condensed phase and condense back into their solid or liquid state.
3. When the rate of evaporation of a solvent in its liquid or solid state becomes equal to the rate of condensation of its gas molecules, no changes take place further.
4. The gas in the container is in equilibrium with its liquid or solid state.

Learn Everything About Raoult’s Law

Factors That Affect Vapor Pressure

The factors that affect vapour pressure are explained in detail below:

1. Surface Area

The solid or liquid surface area in contact with the gas does not affect the vapour pressure. This is because vapour pressure is an intrinsic molecular property. 

2. Types of Molecules

The types of molecules that make up a solid or liquid determine its vapour pressure. If the intermolecular forces between molecules are:

1. Relatively strong, the vapour pressure will be relatively low.
2. Relatively weak, the vapour pressure will be relatively high.

For example- The vapour pressure of diethyl ether \({{\rm{C}}_{\rm{4}}}{{\rm{H}}_{{\rm{10}}}}{\rm{O}}\) at \({\rm{25}}{\,^{\rm{o}}}{\rm{C}}\) is found to be \(520\) torrs, and that of ethyl alcohol \({{\rm{C}}_2}{{\rm{H}}_6}{\rm{O}}\) at \({\rm{25}}{\,^{\rm{o}}}{\rm{C}}\) is found to be \(75\) torrs. The high vapour pressure in diethyl ether is due to the relatively weak dipole-dipole forces and London dispersion forces between its molecules. Strong hydrogen bonding interactions also occur in ethyl alcohol, apart from dipole-dipole forces and London dispersion forces. This accounts for the relatively strong bonding between its molecules that results in low vapour pressure. Due to strong intermolecular forces, fewer molecules move into the vapour phase, thereby lowering the vapour pressure.

3. Temperature

By increasing the temperature, more molecules will have enough thermal energy to escape from the liquid or solid state to the gaseous state. Few molecules have sufficient energy to escape from the liquid or solid state to the gaseous state at a lower temperature.

The variation of the vapour pressure of a liquid with its temperature is graphically represented below. The line on the graph represents the boiling temperature of water.

The vapour pressure of a substance is directly proportional to its temperature. As temperature increases, vapour pressure also increases.

Lowering of Vapour Pressure

We know that the vapour pressure of a liquid is determined by the ease with which its molecules are able to escape the surface of the liquid and enter the gaseous phase.

A liquid evaporating easily will have a relatively large number of its molecules in the gas phase, exerting more vapour pressure over the surface of the liquid at equilibrium conditions. Liquids that do not evaporate easily have less number of their molecules in the gaseous phase and will exert a lower vapour pressure over the surface of the liquid. 

Relative Lowering of Vapour Pressure

On adding a nonvolatile solute to a solvent, the vapour pressure of the solution (solute + solvent) becomes lower than the vapour pressure above the pure solvent.

The Figure above shows the surface of a pure solvent compared to a solution (solute + solvent). 

In the picture on the left, the surface is entirely occupied by solvent molecules, some of which will evaporate and form a vapour. On the right, a nonvolatile solute has been dissolved into the solvent. Nonvolatile means that the solute itself has little tendency to evaporate. 

As some of the surfaces of the solution are now occupied by solute particles, there is less room for solvent molecules to escape the liquid state and enter the gaseous state. This results in less vapour pressure over the surface of the solution. Thus, the addition of nonvolatile solute results in a lowering of the vapour pressure of the solvent.

1. The lowering of the vapour pressure depends on the number of solute particles that have been dissolved. Hence, the relative lowering of vapour pressure is a colligative property of the solution. The only requirement is that the solute and solvent dissolved should not undergo any type of chemical reaction.
2. While the chemical nature of the solute is not a factor, it is important to take into account whether the solute is an electrolyte or a nonelectrolyte. This is because ionic compounds are strong electrolytes and dissociate into their constituent ions when they dissolve in a solvent. The dissociation of the ionic compound increases the number of dissolved particles. 

Vapour Pressure Lowering as a Colligative property

Vapour pressure is not a colligative property. This is because vapour pressure does not depend on the number of solute particles but depends on the nature of the solute. 

Relative vapour pressure is a colligative property, i.e. it depends on the number of solute particles. The dissociation of ionic compounds will affect the vapour pressure of the solution. With an increase in the number of dissociated solute ions, the vapour pressure of the solution decreases.

For example, it was found that the lowering of vapour pressure of a solution consisting of \({\rm{1}}\,{\rm{M}}\,{\rm{NaCl}}\) is more than that of a solution having \({\rm{1}}\,{\rm{M}}\) glucose. The concentration of both solutions at standard temperature and pressure is the same. The reason behind this difference can be accounted for by the following reactions.

\({\rm{NaCl(s)}} \to {\rm{N}}{{\rm{a}}^{\rm{ + }}}{\rm{(aq) + C}}{{\rm{l}}^{\rm{ – }}}{\rm{(aq)}} \Rightarrow {\rm{2}}\) dissolved particles 

\({{\rm{C}}_6}{{\rm{H}}_{12}}{{\rm{O}}_6}(\;{\rm{s}}) \to{{\rm{C}}_6}{{\rm{H}}_{12}}{{\rm{O}}_6}{\rm{(aq)}} \Rightarrow 1\) dissolved particles

Sodium chloride is an ionic compound that dissociates into two ions, whereas glucose being an organic compound does not dissociate. Therefore, equal concentrations of sodium chloride and glucose solution will result in twice as many dissolved particles in the case of sodium chloride. The vapour pressure of the sodium chloride solution will be lowered twice the amount of the glucose solution.

Relative Lowering of Vapour Pressure Formula

Experimentally, we know that the vapour pressure of the solvent above a solution containing a nonvolatile solute (i.e., a solute that does not have a vapour pressure of its own) is directly proportional to the mole fraction of solvent in the solution. This behaviour is summed up in Raoult’s Law:

\({{\rm{P}}_{{\rm{solution}}}}\,{\rm{ = }}\,{{\rm{X}}_{{\rm{solvent}}}}\, \times \,{{\rm{P}}^{\rm{o}}}_{{\rm{solvent}}}\)

Where,

\({{\rm{P}}_{{\rm{solution }}}}\) is the vapour pressure of the solution

\({{\rm{X}}_{{\rm{solvent }}}}\) is the mole fraction of the solvent

\({{\rm{P}}^{\rm{o}}}_{{\rm{solvent}}}\) is the vapour pressure of the pure solvent.

In a binary solution, if a volatile solute is added to the volatile solvent, each solute’s and solvent’s component is added to the total pressure

\({{\rm{P}}_{{\rm{solution}}}}\,{\rm{ = }}\,\left( {{{\rm{X}}_{{\rm{solvent}}}} \times {{\rm{P}}^{\rm{o}}}_{{\rm{solvent}}}} \right)\,{\rm{ + }}\,\left( {{{\rm{X}}_{{\rm{solute}}}} \times {{\rm{P}}^{\rm{o}}}_{{\rm{solute}}}} \right)\)

Note that

i. When \({{\rm{X}}_{{\rm{solvent }}}} = 1\)  (pure solvent), \({{\rm{P}}_{{\rm{solution }}}} = {{\rm{P}}^{\rm{o}}}_{{\rm{solvent}}}\), and

ii. When \({{\rm{X}}_{{\rm{solvent }}}} < 1\) (solute(s) present),  \({{\rm{P}}_{{\rm{solution }}}}{\rm{ <  }}{{\rm{P}}^{\rm{o}}}_{{\rm{solvent}}}\) (i.e., the vapour pressure of the solvent above the solution is lower than the vapour pressure above the pure solvent).

The following graph shows the vapour pressure for water (solvent) at \({\rm{90}}{\,^{\rm{o}}}{\rm{C}}\) as a function of the mole fraction of water in several solutions containing sucrose (a nonvolatile solute). 

The vapour pressure of water decreases as the concentration of sucrose increases.

According to Raoult’s Law, the partial vapour pressure of two components, solvent \(\left( {\rm{A}} \right)\) and solute \(\left( {\rm{B}} \right),\) of a solution may be given as:

\({{\rm{P}}_{\rm{A}}}{\rm{ = }}\,{{\rm{X}}_{\rm{A}}} \times {{\rm{P}}^{\rm{o}}}_{\rm{A}}\)…..eq \(\left( 1 \right)\)

\({{\rm{P}}_{\rm{B}}}\,{\rm{ = }}\,{{\rm{X}}_{\rm{B}}} \times {{\rm{P}}^{\rm{o}}}_{\rm{B}}\)

Total pressure \({\rm{P}}\) is given by-

\({\rm{P}}{\mkern 1mu} \,{\rm{ = }}\,{\mkern 1mu} {{\rm{P}}_{\rm{A}}}{\rm{ + }}{{\rm{P}}_{\rm{B}}}\)

\({\rm{P}}\,{\rm{ = }}\,{{\rm{X}}_{\rm{A}}} \times {{\rm{P}}^{\rm{o}}}_{\rm{A}}{\rm{ + }}{{\rm{X}}_{\rm{B}}} \times {{\rm{P}}^{\rm{o}}}_{\rm{B}}\)

Where \({{\rm{P}}^{\rm{o}}}_{\rm{A}}\) and \({{\rm{P}}^{\rm{o}}}_{\rm{B}}\) are respective vapour pressures in pure form and \({{\rm{X}}_{\rm{A}}}\) and \({{\rm{X}}_{\rm{B}}}\) are respective mole fractions of components \({\rm{A}}\) and \({\rm{B}}.\)

The decrease in vapour pressure of the solvent \(\left( {{\rm{\Delta }}{{\rm{P}}_{\rm{A}}}} \right)\)  is given by-

\(\Delta {{\rm{P}}_{\rm{A}}}{\mkern 1mu} {\rm{ = }}\,{\mkern 1mu} {{\rm{P}}^{\rm{o}}}_{\rm{A}}{\mkern 1mu} {\rm{ – }}{\mkern 1mu} {{\rm{P}}_{\rm{A}}}\)………..eq \(\left( 2 \right)\)

Substituting eq \(\left( 1 \right)\) in eq \(\left( 2 \right),\) we get-

\(\Delta {{\rm{P}}_{\rm{A}}}{\mkern 1mu} {\rm{ = }}{\mkern 1mu} {{\rm{P}}^{\rm{o}}}_{\rm{A}}{\mkern 1mu} {\rm{ – }}{\mkern 1mu} {{\rm{X}}_{\rm{A}}}{{\rm{P}}^{\rm{o}}}_{\rm{A}}\)

\(\Delta {{\rm{P}}_{\rm{A}}}\,{\rm{ = }}\,{{\rm{P}}^{\rm{o}}}_{\rm{A}}\left( {{\rm{1 – }}{{\rm{X}}_{\rm{A}}}} \right)\)…….eq \(\left( 3 \right).\)

We know the total mole fraction of a solution is equal to \(1\). So, 

\({{\rm{X}}_{\rm{A}}}{\rm{ + }}{{\rm{X}}_{\rm{B}}}\,{\rm{ = }}\,{\rm{1}}\)

\({{\rm{X}}_{\rm{B}}}\,{\rm{ = }}\,{\rm{1}}\,{\rm{ – }}\,{{\rm{X}}_{\rm{A}}}\)…………eq \(\left( 4 \right)\)

Substituting eq \(\left( 4 \right)\) in eq \(\left( 3 \right),\) the decrease in vapour pressure of the solvent \(\left( {{\rm{\Delta }}{{\rm{P}}_{\rm{A}}}} \right)\) is given by-

\(\Delta {{\rm{P}}_{\rm{A}}}\,{\rm{ = }}\,{{\rm{P}}^{\rm{o}}}{{\rm{X}}_{\rm{B}}}\)

Using eq \(\left( 2 \right)\) in the above equation, we get,

\({{\rm{P}}^{\rm{o}}}_{\rm{A}}\,{\rm{ – }}\,{{\rm{P}}_{\rm{A}}}\,{\rm{ = }}\,{{\rm{P}}^{\rm{o}}}_{\rm{A}}{{\rm{X}}_{\rm{B}}}\)

\(\Rightarrow \,\frac{{{{\rm{P}}^{\rm{o}}}_{\rm{A}}{\rm{ – }}{{\rm{P}}_{\rm{A}}}}}{{{{\rm{P}}^{\rm{o}}}_{\rm{A}}}}\,{\rm{ = }}\,{{\rm{X}}_{\rm{B}}}\)

\(\Rightarrow {\mkern 1mu} \frac{{{{\rm{P}}^{\rm{o}}}_{\rm{A}}{\rm{ – }}{{\rm{P}}_{\rm{A}}}}}{{{{\rm{P}}^{\rm{o}}}_{\rm{A}}}}{\mkern 1mu} {\rm{ = }}{\mkern 1mu} \frac{{{{\rm{n}}_{\rm{B}}}}}{{{{\rm{n}}_{\rm{A}}}{\rm{ + }}{{\rm{n}}_{\rm{B}}}}}\)

Where, \({{\rm{n}}_{\rm{B}}}\) is the number of moles of the solute and \({{\rm{n}}_{\rm{A}}}\) is the number of moles of the solvent.

For dilute solutions, \({{\rm{n}}_{\rm{A}}} >  > {{\rm{n}}_{\rm{B}}},\) hence, \({{\rm{n}}_{\rm{A}}}{\rm{ + }}{{\rm{n}}_{\rm{B}}} \simeq {{\rm{n}}_{\rm{A}}}\)  

The above equation can be written as-

\(\frac{{{{\rm{P}}^{\rm{o}}}_{\rm{A}}{\rm{ – }}{{\rm{P}}_{\rm{A}}}}}{{{{\rm{P}}^{\rm{o}}}_{\rm{A}}}}{\rm{ = }}\frac{{{{\rm{n}}_{\rm{B}}}}}{{{{\rm{n}}_{\rm{A}}}}}\)

Consequences of Vapour Pressure Lowering

The following consequences are observed due to vapour pressure lowering-

1. Elevation in boiling point 

Salts are often added to foods that require boiling. This is because salt being a nonvolatile solute, increases the boiling point of water.

The normal boiling point is the temperature at which the vapour pressure of a liquid becomes equal to atmospheric pressure. If the intermolecular forces are small, the liquid has a high vapour pressure. Little heat energy will have to be added to separate the molecules so that the boiling point will be low.

Adding a nonvolatile solute to a pure solvent lowers the vapour pressure of the solution. Due to this lowering of vapour pressure, more heat must be supplied to the solution to bring its vapour pressure up to the atmospheric pressure. The boiling point elevation is the difference in temperature between the boiling point of the solution to that of the pure solvent. \(\Delta {\rm{Tb}}\) represents the boiling point elevation in the graph below.

2. Depression inzing Point 

In cold countries, salt is added to clear icy roads making driving easier. This is also known as the salting of roads. By adding salt, the melting point of the ice is decreased, and the ice melts more quickly.This happens because of the lowering of vapour pressure.

Adding a nonvolatile solute to a pure solvent lowers the vapour pressure of the solution. The lowering of the vapour pressure of the solution results in a decrease in thezing point of the solution compared to the solvent. The difference in temperature between thezing point of the pure solvent and that of the solution gives thezing point depression. On the graph, thezing point depression is represented by \({\rm{\Delta Tf}}.\)

The vapour pressure of a solution (blue) is lower than the vapour pressure of a pure solvent (purple). As a result, thezing point of a solution decreases when any solute is dissolved into a solvent.

Vapour Pressure Lowering Example Problems With Solutions

Q.1: At \({\rm{2}}{{\rm{5}}^{\rm{o}}}{\rm{C}}\) the vapor pressure of pure benzene is \(93.9\) torr. When a nonvolatile solute is added to benzene, the vapour pressure of benzene is lowered to \(91.5\) torr. Calculate the mole fraction of the solute and the solvent?
Ans:
Given \({{\rm{P}}^{\rm{o}}}_{{\rm{benzene}}}{\rm{ = }}\,{\mkern 1mu} {\rm{93}}.{\rm{9}}\,{\rm{torr}},\,{\mkern 1mu} {{\rm{P}}_{{\rm{benzene}}}}\,{\rm{ = }}\,{\rm{91}}.{\rm{5}}\,{\rm{torr}}\)
Vapor pressure lowering, \(\Delta {\rm{P}}\,{\mkern 1mu} {\rm{ = }}\,{\mkern 1mu} {{\rm{P}}^{\rm{o}}}_{{\rm{benzene}}}{\mkern 1mu} {\rm{ – }}{\mkern 1mu} {{\rm{P}}_{{\rm{benzene}}}}\,{\rm{ = }}\,{\mkern 1mu} {\left( {{\rm{93}}.{\rm{9}}{\mkern 1mu} {\rm{ – }}{\mkern 1mu} {\rm{91}}.{\rm{5}}} \right)_{{\mkern 1mu} {\rm{torr}}}}\)
\({\rm{ = }}\,{\rm{2}}.{\rm{4}}\,{\rm{torr}}\)
From Raoult’s Law we know that, \(\Delta {\rm{P}}{\mkern 1mu} {\rm{ = }}{\mkern 1mu} {{\rm{P}}^{\rm{o}}}_{{\rm{benzene}}}{{\rm{X}}_{{\rm{solute}}}} \Rightarrow {\mkern 1mu} {{\rm{X}}_{{\rm{solute}}}}{\mkern 1mu} {\rm{ = }}{\mkern 1mu} \frac{{\Delta {\rm{P}}}}{{{{\rm{P}}^{\rm{o}}}_{{\rm{benzene}}}}}\)
\({{\rm{X}}_{{\rm{solute}}}}{\mkern 1mu} {\rm{ = }}{\mkern 1mu} \frac{{{\rm{2}}.{\rm{4}}}}{{{\rm{93}}.{\rm{9}}}}{\mkern 1mu} {\rm{ = }}{\mkern 1mu} {\rm{0}}.{\rm{026}}\)
\({{\rm{X}}_{{\rm{solvent}}}}{\mkern 1mu} {\rm{ = }}{\mkern 1mu} \,{\rm{1}}{\mkern 1mu} {\rm{ – }}{\mkern 1mu} {{\rm{X}}_{{\rm{solute}}{\mkern 1mu} }}{\rm{ = }}{\mkern 1mu} {\rm{1}}\,{\mkern 1mu} {\rm{ – }}{\mkern 1mu} \,{\rm{0}}.{\rm{026}}{\mkern 1mu} \,{\rm{ = }}\,{\mkern 1mu} {\rm{0}}.{\rm{974}}\)

Q.2: The vapour pressure of a pure liquid at \({\rm{298K}}\) is \({\rm{4 \times 1}}{{\rm{0}}^{\rm{4}}}\frac{{\rm{N}}}{{{{\rm{m}}^{\rm{2}}}}}\). On adding a nonvolatile solute, the vapour pressure of the solution becomes \({\rm{3}}{\rm{.65 \times 1}}{{\rm{0}}^{\rm{4}}}\frac{{\rm{N}}}{{{{\rm{m}}^{\rm{2}}}}}\). Calculate the relative vapour pressure, lowering of vapour pressure and relative lowering of vapour pressure.
Ans:
Given, the vapour pressure of pure liquid, \({{\rm{P}}^{\rm{o}}}{\rm{ = 4 \times 1}}{{\rm{0}}^{\rm{4}}}\frac{{\rm{N}}}{{{{\rm{m}}^{\rm{2}}}}}\)  
Vapour pressure of solution, \({\rm{P = 3}}{\rm{.65 \times 1}}{{\rm{0}}^{\rm{4}}}\frac{{\rm{N}}}{{{{\rm{m}}^{\rm{2}}}}}\)
Temperature, \({\rm{T = 298\;K}}\)
Relative vapour pressure, \(\frac{{\rm{P}}}{{{{\rm{P}}^ \circ }}}{\rm{ = }}\frac{{{\rm{3}}{\rm{.65 \times 1}}{{\rm{0}}^{\rm{4}}}\frac{{\rm{N}}}{{{{\rm{m}}^{\rm{2}}}}}}}{{{\rm{4 \times 1}}{{\rm{0}}^{\rm{4}}}\frac{{\rm{N}}}{{{{\rm{m}}^{\rm{2}}}}}}}{\rm{ = 0}}{\rm{.9125}}\)
Lowering of vapour pressure, \({{\rm{P}}^{\rm{o}}}{\rm{ – P = 4 \times 1}}{{\rm{0}}^{\rm{4}}}\frac{{\rm{N}}}{{{{\rm{m}}^{\rm{2}}}}}{\rm{ – 3}}{\rm{.65 \times 1}}{{\rm{0}}^{\rm{4}}}\frac{{\rm{N}}}{{{{\rm{m}}^{\rm{2}}}}}\)
\({\rm{ = 0}}{\rm{.35 \times 1}}{{\rm{0}}^{\rm{4}}}\frac{{\rm{N}}}{{{{\rm{m}}^{\rm{2}}}}}\)
Relative lowering of pressure, \(\frac{{{{\rm{P}}^{\rm{o}}}{\rm{ – P}}}}{{{{\rm{P}}^{\rm{o}}}}}{\rm{ = }}\frac{{{\rm{0}}{\rm{.35 \times 1}}{{\rm{0}}^{\rm{4}}}\frac{{\rm{N}}}{{{{\rm{m}}^{\rm{2}}}}}}}{{{\rm{4 \times 1}}{{\rm{0}}^{\rm{4}}}\frac{{\rm{N}}}{{{{\rm{m}}^{\rm{2}}}}}}}{\rm{ = 0}}{\rm{.0875}}\)

Summary

The properties of a solution are dependent on the amount of solute added to it. Vapour pressure is a vital property of the solution but is not a colligative property of the solution. Whereas, relative lowering of vapour pressure depends on the number of solutes added and is a colligative property of the solution. The elevation in boiling point, depression in thezing point of a substance are all consequences of vapour pressure lowering. Through this article, we learnt how the addition of a solute lowers the vapour pressure of the solvent. We also learnt how the vapour pressure is related to the mole fraction of the solute and consequently to its molar mass. The various real-life evidence of vapour pressure lowering is also covered in this article.

FAQs on Vapour Pressure Lowering

The most frequently asked questions about Lowering of Vapour Pressure are answered here:

Q.1: What is the lowering of the Vapour pressure of a solution? Ans: The phenomenon in which the vapour of a solvent (after the addition of a nonvolatile solute) has a value less than the vapour pressure of the same solvent in its pure form is called lowering of vapour pressure.

Q.2: What is lowering and relative lowering of Vapour pressure? Ans: On adding a nonvolatile solute to a pure solvent, the vapour pressure of the solvent found is lower than the vapour pressure of the solvent in its pure form. This phenomenon is called lowering of vapour pressure. In a solution consisting of solute \({\rm{B}}\) and solvent \({\rm{A}},\) \({{\rm{P}}^{\rm{o}}}_{\rm{A}}\) and \({{\rm{P}}^{\rm{o}}}_{\rm{B}}\) as respective vapour pressures of solvent and solute in pure form, and  \({{\rm{X}}_{\rm{A}}}\) and \({{\rm{X}}_{\rm{B}}}\) as their respective mole fractions. The lowering of vapour pressure is given by- \(\Delta {{\rm{P}}_{\rm{A}}}\,{\rm{ = }}\,{{\rm{P}}^{\rm{o}}}_{\rm{B}}{{\rm{X}}_{\rm{B}}}\) The relative lowering of vapour is the ratio of lowering of vapour pressure to the vapour pressure of the pure solvent. This is represented by the following equation- \(\frac{{{{\rm{P}}^{\rm{o}}}_{\rm{A}}{\rm{ -}}{{\rm{P}}_{\rm{A}}}}}{{{{\rm{P}}^{\rm{o}}}_{\rm{A}}}}{\rm{ = }}{{\rm{X}}_{\rm{B}}}\)

Q.3: What is Raoult’s Law of lowering of Vapour pressure? Ans: Raoult’s law states that the vapour pressure of the solvent above a solution containing a nonvolatile solute is directly proportional to the mole fraction of the solvent in the solution.  \({{\rm{P}}_{{\rm{solution }}}}{\rm{ = }}{{\rm{X}}_{{\rm{solvent }}}}{\rm{ \times }}{{\rm{P}}^{\rm{o}}}_{{\rm{solvent}}}\) Where, \({{\rm{P}}_{{\rm{solution }}}}\) is the vapour pressure of the solution \({{\rm{X}}_{{\rm{solvent}}}}\) is the mole fraction of the solvent \({{\rm{P}}^{\rm{o}}}_{{\rm{solvent}}}\) is the vapour pressure of the pure solvent

Q.4: Why vapour pressure lowering is not a Colligative property? Ans: Colligative properties of the solution are those which do not depend upon the nature of solute but only on the quantity of solute in a solution. As lowering of vapour pressure depends upon the nature of solute rather than quantity, it is not a colligative property. But the relative lowering of vapour pressure is a colligative property because it depends on the quantity of solute rather than its nature.

Q.5: How is Vapour pressure-lowering related to a rise in the boiling point of the solution? Ans: The normal boiling point is the temperature at which the vapour pressure of a liquid becomes equal to atmospheric pressure. If the intermolecular forces are small, the liquid has a high vapour pressure. Little heat energy will have to be able to separate the molecules, and that will have higher vapour pressure, so the boiling point will be low. On adding a nonvolatile solute to a solvent in its pure form, the vapour pressure of the solvent in the solution is lower than the vapour pressure of the solvent in its pure form. As the vapour of the solvent in the solution is less, more heat energy will be required to bring the solution to its boiling point. Hence, vapour pressure lowering leads to an elevation in the boiling point.

Learn Relative Lowering of Vapour Pressure of a Solution Here

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