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On adding a non volatile solute to a solvent, the vapour pressure of solvent decreases and becomes z ×(vapour pressure of solvent) where z is ?
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On adding a non volatile solute to a solvent, the vapour pressure of s...
Effect of Non-Volatile Solute on Vapour Pressure of Solvent


When a non-volatile solute is added to a solvent, the vapour pressure of the solvent decreases. The decrease in vapour pressure can be explained using Raoult's law.


Raoult's Law


Raoult's law states that the vapour pressure of a solution is directly proportional to the mole fraction of the solvent in the solution. Mathematically, it can be expressed as:


vapour pressure of solution = mole fraction of solvent x vapour pressure of pure solvent


Explanation


When a non-volatile solute is added to a solvent, the number of solvent molecules at the surface of the solution decreases. This is because the solute molecules occupy some of the space at the surface of the solution. As a result, the rate of evaporation of solvent molecules decreases, leading to a decrease in the vapour pressure of the solvent.


Colligative Property


The decrease in vapour pressure of the solvent is an example of a colligative property. Colligative properties depend only on the number of solute particles in the solution, not on the identity of the solute particles. Other examples of colligative properties include boiling point elevation, freezing point depression, and osmotic pressure.


Vapour Pressure Depression


The decrease in vapour pressure of the solvent can be expressed as:


vapour pressure of solution = mole fraction of solvent x vapour pressure of pure solvent


Since the solute is non-volatile, its contribution to the vapour pressure of the solution is negligible. Therefore, the vapour pressure of the solution is mainly determined by the mole fraction of the solvent. The decrease in vapour pressure can be expressed as:


ΔP = P0 - P


Where ΔP is the vapour pressure depression, P0 is the vapour pressure of the pure solvent, and P is the vapour pressure of the solution.


From Raoult's law, we know that:


P = Xs x P0


Where Xs is the mole fraction of the solvent in the solution. Substituting this expression for P in the equation for ΔP, we get:


ΔP = P0 - Xs x P0 = P0 (1 - Xs)


Therefore, z in the given expression is (1 - Xs).
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On adding a non volatile solute to a solvent, the vapour pressure of s...
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Direction: Read the passage given below and answer the following questions: Boiling point or freezing point of liquid solution would be affected by the dissolved solids in the liquid phase. A soluble solid in solution has the effect of raising its boiling point and depressing its freezing point. The addition of non-volatile substances to a solvent decreases the vapour pressure and the added solute particles affect the formation of pure solvent crystals. According to many researches the decrease in freezing point directly correlated to the concentration of solutes dissolved in the solvent. This phenomenon is expressed as freezing point depression and it is useful for several applications such as freeze concentration of liquid food and to find the molar mass of an unknown solute in the solution. Freeze concentration is a high quality liquid food concentration method where water is removed by forming ice crystals. This is done by cooling the liquid food below the freezing point of the solution. The freezing point depression is referred as a colligative property and it is proportional to the molar concentration of the solution (m), along with vapour pressure lowering, boiling point elevation, and osmotic pressure. These are physical characteristics of solutions that depend only on the identity of the solvent and the concentration of the solute. The characters are not depending on the solute’s identity. Q.When a non volatile solid is added to pure water it will

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On adding a non volatile solute to a solvent, the vapour pressure of solvent decreases and becomes z ×(vapour pressure of solvent) where z is ?
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