Raoult's Law


The vapor pressure of a solution is lower than that of a pure solvent. Raoult’s Law relates the vapor pressure, P, of a solution, the solute concentration (mole fraction: χ) and the vapor pressure of the pure solvent P°. Raoult’s Law is applicable to ideal and dilute solutions.

For solutions containing volatile (non-electrolyte) solutes, the vapor pressure of each component in the solution (i) is proportional to its concentration in the solution such that

\[P_{i} = \chi_{i} P^{\circ}_{i}\] Each component (i) has a lower vapor pressure than its pure substance form. The lower the concentration of the component, the lower its vapor pressure in the solution.

Adding the vapor pressures together would give the total vapor pressure of the solution such that

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

For example, an aqueous solution containing dissolved benzene is a two-component solution where the solvent (water) is volatile and the solute (benzene) is a volatile non-electrolyte. Therefore, i = water and benzene and the vapor pressure of both would be determined (using the equation above).

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


Practice Problems


Problem 1

What is the vapor pressure of a solution containing 1 mol ClF and 1 mol Cl2? The vapor pressure of pure ClF at 200 K is 242 kPa and the vapor pressure of pure Cl2 at 200 K is 12.3 kPa.

Solution

Get moles of mixture.

\[\begin{align*} n_{\mathrm{mixture}} &= n_{\mathrm{ClF}} + n_{\mathrm{Cl_2}} \\[1.5ex] &= 1 + 1 \\[1.5ex] &= 2~\mathrm{mol} \end{align*}\]

Get mole fraction of each solute.

\[\begin{align*} \chi_{\mathrm{ClF}} &= \dfrac{n_{\mathrm{ClF}}}{n_{\mathrm{mixture}}}\\[1.5ex] &= \dfrac{1~\mathrm{mol~ClF}}{2~\mathrm{mol~mixture}} \\[1.5ex] &= 0.5 \\[4ex] \chi_{\mathrm{Cl_2}} &= \dfrac{n_{\mathrm{Cl_2}}}{n_{\mathrm{mixture}}}\\[1.5ex] &= \dfrac{1~\mathrm{mol~Cl_2}}{2~\mathrm{mol~mixture}} \\[1.5ex] &= 0.5 \end{align*}\]

Get vapor pressure of each solute in the mixture.

\[\begin{align*} P_{\mathrm{ClF}} &= \chi_{\mathrm{ClF}} \times P^{\circ}_{\mathrm{ClF}} \\[1.5ex] &= 0.5 \times 242~\mathrm{kPa} \\[1.5ex] &= 121~\mathrm{kPa} \\[4ex] P_{\mathrm{Cl_2}} &= \chi_{\mathrm{Cl_2}} \times P^{\circ}_{\mathrm{Cl_2}} \\[1.5ex] &= 0.625 \times 12.3~\mathrm{kPa} \\[1.5ex] &= 6.15~\mathrm{kPa} \end{align*}\]

Get vapor pressure of the mixture.

\[\begin{align*} P_{\mathrm{mixture}} &= P_{\mathrm{ClF}} + P_{\mathrm{Cl_2}} \\[1.5ex] &= 121~\mathrm{kPa} + \mathrm{6.15}~\mathrm{kPa}\\[1.5ex] &= 127.15~\mathrm{kPa} \end{align*}\]


Problem 2

What is the vapor pressure of a solution containing 0.30 mol ClF and 0.50 mol Cl2? The vapor pressure of pure ClF at 200 K is 242 kPa and the vapor pressure of pure Cl2 at 200 K is 12.3 kPa.

Solution

Get moles of mixture.

\[\begin{align*} n_{\mathrm{mixture}} &= n_{\mathrm{ClF}} + n_{\mathrm{Cl_2}} \\[1.5ex] &= 0.30 + 0.5 \\[1.5ex] &= 0.80~\mathrm{mol} \end{align*}\]

Get mole fraction of each solute.

\[\begin{align*} \chi_{\mathrm{ClF}} &= \dfrac{n_{\mathrm{ClF}}}{n_{\mathrm{mixture}}}\\[1.5ex] &= \dfrac{0.30~\mathrm{mol~ClF}}{0.80~\mathrm{mol~mixture}} \\[1.5ex] &= 0.375 \\[4ex] \chi_{\mathrm{Cl_2}} &= \dfrac{n_{\mathrm{Cl_2}}}{n_{\mathrm{mixture}}}\\[1.5ex] &= \dfrac{0.50~\mathrm{mol~Cl_2}}{0.80~\mathrm{mol~mixture}} \\[1.5ex] &= 0.625 \end{align*}\]

Get vapor pressure of each solute in the mixture.

\[\begin{align*} P_{\mathrm{ClF}} &= \chi_{\mathrm{ClF}} \times P^{\circ}_{\mathrm{ClF}} \\[1.5ex] &= 0.375 \times 242~\mathrm{kPa} \\[1.5ex] &= 90.75~\mathrm{kPa} \\[4ex] P_{\mathrm{Cl_2}} &= \chi_{\mathrm{Cl_2}} \times P^{\circ}_{\mathrm{Cl_2}} \\[1.5ex] &= 0.625 \times 12.3~\mathrm{kPa} \\[1.5ex] &= 7.688~\mathrm{kPa} \end{align*}\]

Get vapor pressure of the mixture.

\[\begin{align*} P_{\mathrm{mixture}} &= P_{\mathrm{ClF}} + P_{\mathrm{Cl_2}} \\[1.5ex] &= 90.75~\mathrm{kPa} + \mathrm{7.688}~\mathrm{kPa}\\[1.5ex] &= 98.438~\mathrm{kPa} \end{align*}\]


Problem 3

50 g of benzene (C6H6) and 45 g of ethanol (C2H6O), both volatile, are added to 350 g of water at 25 °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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