# 6.4 Colligative properties  (Page 3/30)

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${P}_{\text{A}}={X}_{\text{A}}{P}_{\text{A}}^{°}$

where P A is the partial pressure exerted by component A in the solution, ${P}_{\text{A}}^{°}$ 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

${P}_{\text{solution}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\sum _{i}{P}_{i}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\sum _{i}{X}_{i}{P}_{i}^{°}$

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:

${P}_{\text{solution}}={X}_{\text{solvent}}{P}_{\text{solvent}}^{°}$

## 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:

${P}_{\text{solution}}={X}_{\text{solvent}}{P}_{\text{solvent}}^{°}$

First, calculate the molar amounts of each solution component using the provided mass data.

$\begin{array}{}\\ \\ 92.1\phantom{\rule{0.2em}{0ex}}\overline{)\text{g}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{3}{\text{H}}_{5}\left(\text{OH}{\right)}_{3}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{1\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{3}{\text{H}}_{5}\left(\text{OH}{\right)}_{3}}{92.094\phantom{\rule{0.2em}{0ex}}\overline{)\text{g}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{3}{\text{H}}_{5}\left(\text{OH}{\right)}_{3}}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}1.00\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{3}{\text{H}}_{5}\left(\text{OH}{\right)}_{3}\\ 184.4\phantom{\rule{0.2em}{0ex}}\overline{)\text{g}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{2}{\text{H}}_{5}\text{OH}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\frac{1\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{2}{\text{H}}_{5}\text{OH}}{46.069\phantom{\rule{0.2em}{0ex}}\overline{)\text{g}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{2}{\text{H}}_{5}\text{OH}}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}4.000\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{2}{\text{H}}_{5}\text{OH}\end{array}$

Next, calculate the mole fraction of the solvent (ethanol) and use Raoult’s law to compute the solution’s vapor pressure.

$\begin{array}{}\\ {X}_{{\text{C}}_{2}{\text{H}}_{5}\text{OH}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{4.000\phantom{\rule{0.2em}{0ex}}\text{mol}}{\left(1.00\phantom{\rule{0.2em}{0ex}}\text{mol}+4.000\phantom{\rule{0.2em}{0ex}}\text{mol}\right)}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}0.800\\ {P}_{\text{solv}}={X}_{\text{solv}}{P}_{\text{solv}}^{°}=0.800\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}0.178\phantom{\rule{0.2em}{0ex}}\text{atm}=0.142\phantom{\rule{0.2em}{0ex}}\text{atm}\end{array}$

## 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?

23.4 torr

## 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:

$\text{Δ}{T}_{\text{b}}={K}_{\text{b}}m$

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

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