Vapor Pressure of a Solution

Vapor Pressure of a Solution

From the illustration above, Initially, two beakers, both containing water solutions of the same solute, are placed under a bell jar. The solution in the left beaker is less concentrated than that in the right beaker, so its vapor pressure is greater.

The partial pressure of vapor in the bell jar is an intermediate value. It is less than the vapor pressure of the solution on the left, but more than that of the solution on the right. As a result, vapor leaves the solution on the left (which becomes more concentrated) and condenses on the solution on the right (which becomes less concentrated). After some time, the two solutions become equal in concentration and in vapor pressure.

During the nineteenth century, chemists observed that the vapor pressure of a volatile solvent was lowered by addition of a nonvolatile solute. Vapor-pressure lowering of a solvent is a colligative property equal to the vapor pressure of the pure solvent minus the vapor pressure of the solution.

 For example, water at 20 C has a vapor pressure of 17.54 mmHg. Ethylene glycol, CH2OHCH2OH, is a liquid whose vapor pressure at 20 C is relatively low; it can be considered to be nonvolatile compared with water. An aqueous solution containing 0.0100 mole fraction of ethylene glycol has a vapor pressure of 17.36 mmHg. Thus, the vapor-pressure lowering, P, of water is Change in   P = 17.54 mmHg -17.36 mmHg= 0.18 mmHg.

In about 1886, the French chemist François Marie Raoult observed that the partial vapor pressure of solvent over a solution of a nonelectrolyte solute depends on the mole fraction of solvent in the solution. Consider a solution of volatile solvent, A, and nonelectrolyte solute, B, which may be volatile or nonvolatile.

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 According to Raoult’s law, the partial pressure of solvent, PA, over a solution equals the vapor pressure of the pure solvent, PA , times the mole fraction of solvent, XA, in the solution.


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If the solute is nonvolatile, PA is the total vapor pressure of the solution. Because the mole fraction of solvent in a solution is always less than 1, the vapor pressure of the solution of a nonvolatile solute is less than that for the pure solvent; the vapor pressure is lowered. In general, Raoult’s law is observed to hold for dilute solutions—that is, solutions in which XA is close to 1. If the solvent and solute are chemically similar, Raoult’s law may hold for all mole fractions. Raoult’s law is displayed graphically 
for two solutions in the figure below….

You can obtain an explicit expression for the vapor-pressure lowering of a solvent in a solution, assuming that Raoult’s law holds and that the solute is a nonvolatile nonelectrolyte. The vapor-pressure lowering, P, is Change inP = P°A- PA

From this equation, you can see that the vapor-pressure lowering is a colligative property— one that depends on the concentration, but not on the nature, of the solute. Thus, if the mole fraction of ethylene glycol, XB, in an aqueous solution is doubled from 0.010 to 0.020, the vapor-pressure lowering is doubled from 0.18 mmHg to 0.36 mmHg. Also, because the previous equation does not depend on the characteristics of the solute (other than its being nonvolatile and a nonelectrolyte), a solution that is 0.010 mole fraction urea, (NH2)2CO, has the same vapor-pressure lowering as
one that is 0.010 mole fraction ethylene glycol. The next example illustrates the use of the previous equation.

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Problem: Calculate the vapor-pressure lowering of water when 5.67 g of glucose, C6H12O6, is dissolved in 25.2 g of water at 25 C. The vapor pressure of water at 25 C is 23.8 mmHg. What is the vapor pressure of the solution?

Problem Strategy: In order to calculate the vapor pressure lowering of the solution described in this
problem, we need to determine the amount that the vapor pressure of the pure solvent is changed ( P
of water) with the addition of a solute (glucose). We can use Raoult’s law to calculate this quantity,
noting that we need to know the mole fraction of the solute (glucose). The vapor pressure of the solution is found by subtracting the vapor-pressure change of the pure solvent from the vapor pressure of the pure solvent.

Solution : This is the glucose solution described in Example 12.4. According to the calculations performed there, the solution is 0.0220 mole fraction glucose.
Therefore, the vapor-pressure lowering is Change in P = PA°XB =23.8 mmHg x 0.0220= 0.524 mmHg
The vapor pressure of the solution is  PA = PA° – Change in P (23.8 – 0.524) mmHg = 23.3 mmHg

An ideal solution of substances A and B is one in which both substances follow Raoult’s law for all values of mole fractions. Such solutions occur when the substances are chemically similar so that the intermolecular forces between A and B molecules are similar to those between two A molecules or between two B molecules.

In this case, we are not restricted to the solute being nonvolatile: both the solute and solvent have significant vapor pressure. Therefore, the total vapor pressure over an ideal solution equals the sum of the partial vapor pressures, each of which is given by Raoult’s law:

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Change in P = PA°XA + PB°XB

Solutions of benzene, C6H6, and toluene, C6H5CH3, are ideal. Note the similarity
in their structures…

Suppose a solution is 0.70 mole fraction benzene and 0.30 mole fraction toluene.
The vapor pressures of pure benzene and pure toluene are 75 mmHg and 22 mmHg,
respectively. Hence, the total vapor pressure is  change in P (75 mmHg x 0.70) + (22 mmHgx 0.30)= 59 mmHg.

Recommended lecture/article: The Valence Bond TheoryAliphatic Hydrocarbons: Definition s and Properties,Effect of Polarity on Molecular Properties

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