${P}_{\text{A}}={X}_{\text{A}}{P}_{\text{A}}^{\xb0}$
where
P
_{A} is the partial pressure exerted by component A in the solution,
${P}_{\text{A}}^{\xb0}$ is the vapor pressure of pure A, and
X
_{A} is the mole fraction of A in the solution. (Mole fraction is a concentration unit introduced in the chapter on gases.)
Recalling that the total pressure of a gaseous mixture is equal to the sum of partial pressures for all its components (Dalton’s law of partial pressures), the total vapor pressure exerted by a solution containing
i components is
${P}_{\text{solution}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}{\displaystyle \sum _{i}{P}_{i}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}{\displaystyle \sum _{i}{X}_{i}{P}_{i}^{\xb0}}$
A nonvolatile substance is one whose vapor pressure is negligible (
P ° ≈ 0), and so the vapor pressure above a solution containing only nonvolatile solutes is due only to the solvent:
${P}_{\text{solution}}={X}_{\text{solvent}}{P}_{\text{solvent}}^{\xb0}$
Calculation of a vapor pressure
Compute the vapor pressure of an ideal solution containing 92.1 g of glycerin, C
_{3} H
_{5} (OH)
_{3} , and 184.4 g of ethanol, C
_{2} H
_{5} OH, at 40 °C. The vapor pressure of pure ethanol is 0.178 atm at 40 °C. Glycerin is essentially nonvolatile at this temperature.
Solution
Since the solvent is the only volatile component of this solution, its vapor pressure may be computed per Raoult’s law as:
${P}_{\text{solution}}={X}_{\text{solvent}}{P}_{\text{solvent}}^{\xb0}$
First, calculate the molar amounts of each solution component using the provided mass data.
$\begin{array}{}\\ \\ 92.1\phantom{\rule{0.2em}{0ex}}\overline{)\text{g}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{3}{\text{H}}_{5}(\text{OH}{)}_{3}}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\frac{1\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{3}{\text{H}}_{5}(\text{OH}{)}_{3}}{92.094\phantom{\rule{0.2em}{0ex}}\overline{)\text{g}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{3}{\text{H}}_{5}(\text{OH}{)}_{3}}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}1.00\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{3}{\text{H}}_{5}(\text{OH}{)}_{3}\\ 184.4\phantom{\rule{0.2em}{0ex}}\overline{)\text{g}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{2}{\text{H}}_{5}\text{OH}}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\frac{1\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{2}{\text{H}}_{5}\text{OH}}{46.069\phantom{\rule{0.2em}{0ex}}\overline{)\text{g}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{2}{\text{H}}_{5}\text{OH}}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}4.000\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{2}{\text{H}}_{5}\text{OH}\end{array}$
Next, calculate the mole fraction of the solvent (ethanol) and use Raoult’s law to compute the solution’s vapor pressure.
$\begin{array}{}\\ {X}_{{\text{C}}_{2}{\text{H}}_{5}\text{OH}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{4.000\phantom{\rule{0.2em}{0ex}}\text{mol}}{\left(1.00\phantom{\rule{0.2em}{0ex}}\text{mol}+4.000\phantom{\rule{0.2em}{0ex}}\text{mol}\right)}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}0.800\\ {P}_{\text{solv}}={X}_{\text{solv}}{P}_{\text{solv}}^{\xb0}=0.800\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}0.178\phantom{\rule{0.2em}{0ex}}\text{atm}=0.142\phantom{\rule{0.2em}{0ex}}\text{atm}\end{array}$
Check your learning
A solution contains 5.00 g of urea, CO(NH
_{2} )
_{2} (a nonvolatile solute) and 0.100 kg of water. If the vapor pressure of pure water at 25 °C is 23.7 torr, what is the vapor pressure of the solution?
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Elevation of the boiling point of a solvent
As described in the chapter on liquids and solids, the
boiling point of a liquid is the temperature at which its vapor pressure is equal to ambient atmospheric pressure. Since the vapor pressure of a solution is lowered due to the presence of nonvolatile solutes, it stands to reason that the solution’s boiling point will subsequently be increased. Compared to pure solvent, a solution, therefore, will require a higher temperature to achieve any given vapor pressure, including one equivalent to that of the surrounding atmosphere. The increase in boiling point observed when nonvolatile solute is dissolved in a solvent, Δ
T
_{b} , is called
boiling point elevation and is directly proportional to the molal concentration of solute species:
$\text{\Delta}{T}_{\text{b}}={K}_{\text{b}}m$
where
K
_{b} is the
boiling point elevation constant , or the
ebullioscopic constant and
m is the molal concentration (molality) of all solute species.
Boiling point elevation constants are characteristic properties that depend on the identity of the solvent. Values of
K
_{b} for several solvents are listed in
[link] .
Boiling Point Elevation and Freezing Point Depression Constants for Several Solvents |
Solvent |
Boiling Point (°C at 1 atm) |
K
_{b} (C
m
^{−1} ) |
Freezing Point (°C at 1 atm) |
K
_{f} (C
m
^{−1} ) |
water |
100.0 |
0.512 |
0.0 |
1.86 |
hydrogen acetate |
118.1 |
3.07 |
16.6 |
3.9 |
benzene |
80.1 |
2.53 |
5.5 |
5.12 |
chloroform |
61.26 |
3.63 |
−63.5 |
4.68 |
nitrobenzene |
210.9 |
5.24 |
5.67 |
8.1 |