When a non volatile solute is dissolved in a pure solvent the vapour pressure of the pure S?

When the amount of nonvolatile solutes is large (typical polymer volume fraction, ϕp>10−1), the contact line remains pinned at its initial position.

From: Droplet Wetting and Evaporation, 2015

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A knowledge of Latin can help us to better understand the English language—specifically, scientific terminology. Consider the word "colligative". Where did that come from? If you know a little Latin, you know that it comes from two Latin words meaning "to tie together".  

A colligative property is a property of a solution that depends only on the number of solute particles dissolved in the solution, and not on their identity. Recall that the vapor pressure of a liquid is determined by how easily its molecules are able to escape the surface of the liquid and enter the gaseous phase. When a liquid evaporates easily, it will have a relatively large number of its molecules in the gas phase and thus will have a high vapor pressure. Liquids that do not evaporate easily have a lower vapor pressure. The figure below shows the surface of a pure solvent compared to a solution. In the picture on the left, the surface is entirely occupied by liquid molecules, some of which will evaporate and form a vapor pressure. On the right, a nonvolatile solute has been dissolved into the solvent. Nonvolatile means that the solute itself has little tendency to evaporate. Because some of the surface is now occupied by solute particles, there is less room for solvent molecules. This results in less solvent being able to evaporate. The addition of a nonvolatile solute results in a lowering of the vapor pressure of the solvent.

When a non volatile solute is dissolved in a pure solvent the vapour pressure of the pure S?
Figure \(\PageIndex{1}\): The solution on the right has had some of its solvent particles replaced by solute particles. Since the solute particles do not evaporate, the vapor pressure of the solution is lower than that of the pure solvent.

The lowering of the vapor pressure depends on the number of solute particles that have been dissolved. The chemical nature of the solute is not important because the vapor pressure is merely a physical property of the solvent. The only requirement is that the solute was only dissolved and that it did not undergo a chemical reaction with the solvent.

While the chemical nature of the solute is not a factor, it is necessary to take into account whether the solute is an electrolyte or a nonelectrolyte. Recall that ionic compounds are strong electrolytes, and thus dissociate into ions when they dissolve. This results in a larger number of dissolved particles. For example, consider two different solutions of equal concentration: one is made from the ionic compound sodium chloride, while the other is made from the molecular compound glucose. The following equations show what happens when these solutions dissolve.

\[\begin{array}{ll} \ce{NaCl} \left( s \right) \rightarrow \ce{Na^+} \left( aq \right) + \ce{Cl^-} \left( aq \right) & 2 \: \text{dissolved particles} \\ \ce{C_6H_{12}O_6} \left( s \right) \rightarrow \ce{C_6H_{12}O_6} \left( aq \right) & 1 \: \text{dissolved particle} \end{array}\nonumber \]

The sodium chloride dissociates into two ions, while the glucose does not dissociate. Therefore, equal concentrations of each solution will result in twice as many dissolved particles in the case of the sodium chloride. The vapor pressure of the sodium chloride solution will be lowered twice the amount of the glucose solution.

Summary

  • A colligative property is a property of a solution that depends only on the number of solute particles dissolved in the solution, and not on their identity.
  • Nonvolatile means that the solute itself has little tendency to evaporate.
  • Addition of a nonvolatile solute to a solution lowers the vapor pressure of the solution.

When a non volatile solute is dissolved in a pure solvent the vapour pressure of the pure S?

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When a non volatile solute is dissolved in a pure solvent the vapour pressure of the pure S?

When a non volatile solute is dissolved in a pure solvent the vapour pressure of the pure S?

Text Solution

`P_(s) = X_(A).P_(B)^(@)` `P_(B)^(@) =P_(s).X_(A)` `P_(S)= X_(B).P_(B)^(@)``P_(B)^(@)= P_(s). X_(B)`

Answer : ( c)

Solution : According to Raoult's law, the vapour pressure of resultant solution is equal to the product of mole fraction and vapour pressure of pure solvent. <br>`P_(s)=X_(8)P_(B)^(0)`