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Relative Lowering of Vapour Pressure: Vapour Pressure, Factors Affecting Vapour Pressure, Relative Lowering of Vapour Pressure, Mathematical Representation, Ostwald-Walker Method, Practice Problems, FAQs

Relative Lowering of Vapour Pressure: Vapour Pressure, Factors Affecting Vapour Pressure, Relative Lowering of Vapour Pressure, Mathematical Representation, Ostwald-Walker Method, Practice Problems, FAQs

Clothes are dried in windy sunlight, preferably than closed indoor rooms. Have you thought about the reasons behind this daily practice?

The drying of clothes is a significant occurrence that employs the concept of vapour pressure.

For example, clothes are typically dried by exposing them to sunshine. The rays of sunlight that land on the damp clothes tend to raise the temperature, which raises the vapour pressure even more. Water vapour moves from a low pressure location (wet clothes) to a high pressure region (surrounding environment). As a result, the clothes dry.

Table of content

  • Relative Lowering of Vapour Pressure
  • Mathematical Representation for Relative Lowering of Vapour Pressure
  • Ostwald-Walker Method
  • Practice Problems
  • Frequently Asked Questions

Relative Lowering of Vapour Pressure

Vapour pressure is the pressure that vapours apply to a liquid when it is in an equilibrium state and a given temperature. As an illustration, consider a pure liquid. On its surface are the molecules of the liquid. Now imagine that a non-volatile solute has been added to this pure liquid. Since the molecules of the solute are not volatile, only the molecules of the solvent (pure liquid) are present in the vapour that rises above the solution. After the solute is added, it is discovered that the solution's vapour pressure, at a particular temperature, is lower than that of the pure liquid.

The molecules of both the pure liquid and the solute are now present on the liquid surface as a result of the solute's addition to the pure liquid (solvent). As a result of a reduction in the number of solvent molecules that escape into the vapour phase, the vapour phase's pressure also decreases.

Relative lowering of vapour pressure refers to the net decrease in vapour pressure of the solvent in the pure solvent and solution states compared to the pure solvent vapour pressure. This decrease in vapour pressure, regardless of its nature, is proportional to the amount of non-volatile solute particles present in the solution and is thus a colligative property.

Mathematical Representation For Relative Lowering of Vapour Pressure

For a solution of a volatile solvent containing a non-volatile solute-

As per Raoult’s Law,

vapour pressure of the solvent

Let, the mole fraction of the solute is and the mole fraction of solvent be

and isvapour pressure of pure solvent.

is the lowering of vapour pressure

and is called relative lowering of vapour pressure.

Mole fraction of solute

Ostwald-Walker method

This provides an experiment to determine how much a solvent's vapour pressure decreases when a non-volatile solute is present. The theory behind this process is that when passing dry air repeatedly through a series of containers containing pure solvent and solution, respectively, the air becomes saturated with the solvent vapours and an equal amount of weight loss occurs in the solution and solvent containers.

In the Ostwald-Walker apparatus, a known quantity of the solution to be examined is present in the first two chambers, and a known quantity of the pure solvent is present in the succeeding two chambers. A U-tube filled with a known quantity of dry anhydrous . is connected to the solvent chamber on one end and kept open at the other end.

Dry air is sent through the solution chamber and passes through the adjacent chambers and U-tube in order before coming out of the U-tube. While doing so, the dry air picks from the solution chambers water vapours proportional to the vapour present above the solution. The wet air further picks up more water vapour from the solvent chambers to get saturated. All the water present in the air is absorbed by the calcium chloride in the U-tube. The increase in weight of the calcium chloride tube should be equal to the weight lost by the solution and solvent chambers.

When the pure solvent chamber is opened to this air, more solvent vapour enters the stream with the air until the solvent's vapour pressure , is reached. Because is greater than , this occurs. As a result, the loss of weight recorded in the solvent chamber is inversely proportional to the amount of .

Loss of weight in the solution chamber

Loss of weight in the solvent chamber

Total weight loss from the solvent and solution chambers

Solvent vapours are absorbed and dry air is released through a U-tube when solvent-saturated air is passed through it. The U-weight tube's gain should be proportional to and equal to the sum of the weights lost by the solution and solvent chambers.

.

Raoult's law can be used to calculate the molecular weight of the solute by using the experimental values.

Practice problems

Q.1.Pure liquid A's vapour pressure is When a non-volatile substance B is added, the solvent's vapour pressure falls to . What is the mole fraction of components A and B in the solution?

Solution:

Q.2. If of glucose is added to of water. The vapour pressure of water() for this aqueous solution is

Solution:Number of moles of glucose

Number of moles of water

Total number of moles

Now, mole fraction of glucose in solution pressure change with respect to initial pressure

i.e.

Vapour pressure of solution

Q.3. Acetone has a vapour pressure of at . The vapour pressure of of a non-volatile substance dissolved in of acetone at was . The substance's molar mass is

Solution:Given

Mass of non-volatile substance

Mass of acetone taken

Molar mass of acetone

Putting the values we get

Q.4.The vapour pressure of of solute inof water at is Calculate the number of moles of solute.

Solution:Given that

Mass of

Molar mass of

Frequently asked questions(FAQs)

Q1.What is a non-volatile substance?
Answer: 
Substance that doesn't easily turn into gas during evaporation. Low vapour pressure and a high boiling point characterise non-volatile compounds. The non-volatile chemicals alcohol, mercury, and gasoline are some examples.

Q2.Give one application of vapour pressure in daily life?
Answer: 
The tea kettle is the best place to use vapour pressure. The moment we pour water into the kettle, it will begin to heat up and produce a rising vapour pressure. The kettle's lid is propelled upward by the pressure that the water vapour puts on it.

Q3. What are the applications of colligative properties in everyday life?
Answer:
In everyday life, we use elevation in boiling point as a colligative characteristic. You may have heard that making a good cup of tea or coffee in the mountains is very difficult. This is due to the fact that air pressure decreases with increasing altitude relative to sea level, causing water to boil at temperatures lower than 373 K. At higher altitudes, water evaporates at a faster rate as a result, producing poorer cups of tea.

Q4.Why is molality used in colligative properties?
Answer:
We are aware that the number of particles in the solution is the only factor affecting the colligative properties. The mass of the solvent, which is unaffected by temperature, determines molality. Molality is therefore used in colligative properties.

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