# Vapor Pressure

The vapor pressure of a solvent is lower than that of a pure liquid. Raoult’s Law relates the vapor pressure, P, of a solution, the solute concentration (mole fraction: $$\chi$$) and the vapor pressure of the pure liquid $$(P^{\circ})$$. Raoult’s Law is applicable to ideal and dilute solutions. To get the vapor pressure of a solution, determine the vapor pressure of the individual solutes and the solvent.

Solvent vapor pressure

$P_{\mathrm{solvent}} = \chi_{\mathrm{solvent}} P^{\circ}_{\mathrm{solvent}}$

If the solutes are volatile, we can determine their vapor pressures as well if we know their mole fraction.

Solute vapor pressure

$P_{\mathrm{solute}} = \chi_{\mathrm{solute}} P^{\circ}_{\mathrm{solute}}$

Note: The vapor pressure of non-volatile substances is essentially zero.

The vapor pressure of a solution can then be determined.

Solution vapor pressure

$P_{\text {soln }}=\sum_{i=1}^{n}\left(P_{\text {solute}}\right)_{i}+P_{\text {solvent}}$

## Practice Problems

### Problem 1

50 g of benzene (C6H6) and 45 g of ethanol (C2H6O), both volatile, are added to 350 g of water at 25 $$^{\circ}$$C. What is the vapor pressure (in atm) of the solvent? What is the vapor pressure (in atm) of the solution? Treat the solution as an ideal solution.

• $$P^{\circ}_{\mathrm{water}}$$ = 0.0313 atm
• $$P^{\circ}_{\mathrm{benzene}}$$ = 0.132 atm
• $$P^{\circ}_{\mathrm{ethanol}}$$ = 0.077 atm
Solution

Solve for moles of each species:

\begin{align*} n_{\mathrm{water}} &= 350~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{18.02~\mathrm{g}} \right ) = 19.4~\mathrm{mol}\\ n_{\mathrm{benzene}} &= 50~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{78.11~\mathrm{g}} \right ) = 0.640~\mathrm{mol}\\ n_{\mathrm{ethanol}} &= 45~\mathrm{g} \left ( \dfrac{\mathrm{mol}}{46.07~\mathrm{g}} \right ) = 0.977~\mathrm{mol}\\[2ex] n_{\mathrm{tot}} &= 19.4 + 0.640 + 0.977 = 21.017~\mathrm{mol~solution} \end{align*}

Solve for mole fraction of each species:

\begin{align*} \chi &= \dfrac{n_i}{n_{\mathrm{tot}}} \\[2ex] \chi_{\mathrm{water}} &= \dfrac{19.4~\mathrm{mol~H_2O}}{21.017~\mathrm{mol~solution}} = 0.923 \\[2ex] \chi_{\mathrm{benzene}} &= \dfrac{0.640~\mathrm{mol~C_6H_6}}{21.017~\mathrm{mol~solution}} = 0.0304 \\[2ex] \chi_{\mathrm{ethanol}} &= \dfrac{0.977~\mathrm{mol~C_2H_6O}}{21.017~\mathrm{mol~solution}} = 0.0465 \end{align*}

Solve for vapor pressure of each (volatile) species:

\begin{align*} P &= \chi P^{\circ}\\[2ex] P_{\mathrm{water}} &= \left ( 0.923 \right ) \left ( 0.0313~\mathrm{atm} \right ) = 0.0289~\mathrm{atm} \\ P_{\mathrm{benzene}} &= \left ( 0.0304 \right ) \left ( 0.132~\mathrm{atm} \right ) = 0.004013~\mathrm{atm} \\ P_{\mathrm{ethanol}} &= \left ( 0.0465 \right ) \left ( 0.077~\mathrm{atm} \right ) = 0.00358~\mathrm{atm} \end{align*}

The vapor pressure of the water is 0.0289 atm which is lower than that of pure water (0.0313 atm).

Find vapor pressure of solution:

\begin{align*} P_{\mathrm{tot}} &= P_{\mathrm{water}} + P_{\mathrm{benzene}} + P_{\mathrm{ethanol}}\\ &= 0.0289 + 0.004013 + 0.00358 = 0.036~\mathrm{atm} \end{align*}

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