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Let’s explore how a nonvolatile solute, a solute that does not escape to the vapor phase, affects vapor pressure. Take a look at a container with two beakers, one containing a solution of hydrochloric acid and another one containing pure water. Over time, the volume of the hydrochloric acid increases and the volume of the water decreases. Why does this happen? The answer lies in terms of vapor pressure. Remember, the total vapor pressure of the container is equal to the equilibrium pressure of gas molecules over the two liquids. The pure solvent has a greater equilibrium vapor pressure than the solution, so its volume would decrease because the water would go to the vapor phase more easily than the solution. The solution increases in volume because it absorbs the water vapor in the container, in an attempt to lower the vapor pressure to reach equilibrium. This experiment indicates that the presence of a nonvolatile solute lowers the vapor pressure of a solvent.

(A) Before

(B) After

(A) Water molecules escape to the vapor phase through evaporation, the solution of hydrochloric acid begins to absorb water molecules in order to reach equilibrium vapor pressure. (B) The beaker containing water is now empty because all the water has evaporated and the volume of the hydrochloric solution increases because water has been absorbed.

Raoult’s Law

A nonvolatile solute lowers the vapor pressure of a solvent because solute particles become surrounded by solvent particles when dissolved. This causes the particles of solvent to evaporate (escape to the vapor phase from the liquid phase) less because it requires more energy for the particles to do this since the solvent to solute attractions are strong. So the vapor pressure of a solution is less than that of the pure solvent it was created with.


Pure Solvent                                   Solution

The pure solvent has a higher vapor pressure than the solution as indicated by the pressure gauge on each container.

Francois M. Raoult discovered this phenomenon and created what is called Raoult’s Law. Raoult’s law states that the vapor pressure of a solution at a particular temperature is equal to the vapor pressure of the pure solvent multiplied by the mole fraction of solvent in the solution.1 In equation form it appears as:

Psol = Xsolv P0solv

Xsolv is the mole fraction of the solvent and P0solvent is the vapor pressure of the pure solvent. This equation is very useful for determining the vapor pressure of a solution.

In addition to solutions containing non-volatile solutes, Raoult’s Law may be used for liquid-liquid solutions containing volatile solutes. Volatile solutes are solutes which escape to the vapor phase at standard conditions. In liquid-liquid solutions containing volatile liquid solutes, each substance, the solute and solvent, expresses its own vapor pressure. Therefore, an altered form of Raoult’s Law is necessary for this case. It is presented here as:

Ptotal = XAP0A + XBP0B + …

The subscripts A and B represent two different volatile liquids in the solution.

Sample Problems

1. Calculate the vapor pressure of the solution when 235 grams of sugar (C12H22O11) are added to 650 mL of water at 40°C. The vapor pressure of water at 40°C is 55 mm Hg.

2. Calculate the vapor pressure of the solution when 55 grams of KI are added to 325 mL of water at 25°C. The vapor pressure of water at 25°C is 23.8 mm Hg.

3. Calculate the vapor pressure of a solution where 35 grams of water is mixed with 165 grams of ethanol, C2H5OH, at 35°C. The vapor pressures of the two substances are 42 mm Hg and 100 mm Hg, respectively.

Answers

1. The molar mass of C12H22O11 is 342 g/mol. By dividing the grams of sugar by the molar mass we obtain the moles:(235g sugar)/ (342g/mol) = 0.687 mol sugarNext we find the moles of water using the grams of water, which is 650 g because the density of water is 1.00 g/mL:

(650 g H2O)/ (18 g/mol) = 36.11 mol H2O

Find the mole fraction of solvent:                

            Xsolv = (mol H2O)/ (mol sugar + mol H2O)

                     = (36.11 mol H2O)/ (0.687 mol sugar + 36.11 mol H2O)

= 0.981

Plug it into the equation for Raoult’s Law and we get:

Psol = (0.981)(55 mm Hg)

= 54.0 mm Hg

The above problem was concerned only with non-electrolyte solutions, or solutions that contained solutes that do not dissociate into ions. In order to figure out the vapor pressure for an electrolyte solution, the process is the same, but you must multiply your moles of solute by the ratio of ions in the solution. For example, if the electrolyte solution was of NaCl, the ratio of ions would be 2, because NaCl dissolves into Na+ and Cl, and you would multiply your moles of solute by that.

2. Find moles of KI:

mol KI = (55g KI)/ (166 g/mol) = 0.3313 mol KI

Now we must multiply that value by 2 because there are two moles of ions for every mole of solid solute.

0.3313 mol KI x 2 = 0.6626 moles particles

Find moles of water: (325 g water)/ (18 g/mol) = 18.06 mol water

Find the mole fraction of water:

Xwater= (18.06 mol water)/ (0.6626 mol particles + 18.06 mol water)

= 0.965

Now we can plug everything into Raoult’s Law to figure out the vapore pressure of solution:

Psol = (0.965)(23.8 mm Hg)

= 23.0 mm Hg

3. Find the moles of water and ethanol:

mol H2O = (35 g H­2O)/ (18 g/mol) = 1.944 mol H2O

mol ethanol= (165 g C2H5OH)/ (46 g/mol) = 3.587 mol C25OH

Now find the mole fractions of each:

XA = (1.944 mol H2O)/ (1.944 mol H2O + 3.587 mol C25OH) = 0.351

XB = (3.587 mol C25OH)/ (1.944 mol H2O + 3.587 mol C25OH) = 0.649

Finally, plug it into the altered form of Raoult’s law to obtain:

Ptotal = (42 mm Hg)(0.351) + (100 mm Hg)(0.649)

= 80. mm Hg

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