Vapour Pressure of Solutions of Solids in Liquids: The vapour pressure of a solution of a solid in a liquid can be explained by Raoult’s law. According to Raoult’s law, the vapour pressure of a solution is directly proportional to the mole fraction of the solvent in the solution. When a non-volatile solute is added to a volatile solvent, the vapour pressure of the solution decreases compared to that of the pure solvent.

Vapour Pressure of Solutions of Solids in Liquids

In the case of a solid-liquid solution, the solid is the solute and the liquid is the solvent. The vapour pressure of the solution is lower than that of the pure solvent because the surface of the liquid is partially occupied by the solute molecules. The presence of solute molecules reduces the number of solvent molecules that can escape into the vapour phase, thereby lowering the vapour pressure of the solution.

The decrease in vapour pressure of the solution can be quantified by the following equation:

ΔP = Xsolute * Psolvent * Kf

Where,

• ΔP is the change in vapour pressure of the solution compared to that of the pure solvent
• Xsolute is the mole fraction of the solute in the solution
• Psolvent is the vapour pressure of the pure solvent
• Kf is the freezing point depression constant

The decrease in vapour pressure due to the addition of a solute can also be used to determine the molecular weight of the solute, using the following equation:

M = Kf * w2 / ΔTf

Where,

• M is the molecular weight of the solute
• Kf is the freezing point depression constant
• w₂ is the weight of the solute added to the solution
• ΔTf is the freezing point depression of the solution compared to that of the pure solvent

In summary, the vapour pressure of a solution of a solid in a liquid is lower than that of the pure solvent, and can be quantified by Raoult’s law. The decrease in vapour pressure can be used to determine the molecular weight of the solute.

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By Team Learning Mantras