Question

The vapor pressure of pure acetone, ${\mathbf{\text{CH}}}_{\mathbf{3}}\mathbf{\text{C}}\mathbf{\left(}\mathbf{\text{O}}\mathbf{\right)}{\mathbf{\text{CH}}}_{\mathbf{3}}$at${\mathbf{30}}^{\mathbf{0}}\mathit{C}$is 0.3270 atm. Suppose 15.0 gof benzophenone, is dissolved inof acetone. Calculate the vapor pressure of acetone above the resulting solution.

Short answer

The vapor pressure of the resulting solution is $0.2985\text{atm}$.

Step-by-step solution

Given data.

The vapor pressure of pure acetone is $0.3270\text{atm}$.

The amount of benzophenone is 15.0 g.

Concept of vapor pressure.

When a liquid and vapor is in equilibrium, the pressure exerted by vapor on the surface of the liquid is known as the vapor pressure. The temperature at which the vapor pressure equals the surrounding pressure is the boiling point of the liquid. The vapor pressure of a pure liquid is always higher than its solution.

Relation between vapor pressure and mole fraction 

The relation between vapor pressure and mole fraction can be expressed as,

$P_1=X_1{P}_1^0$ ……(i)

where P₁ is the vapor pressure of the solution, X₁is the mole fraction of the solute,$P_1^0$ is the vapor pressure of the pure solvent.

Calculation of n1 and n2

${\text{n}}_1\text{=} \text{number} \text{of} \text{moles} \text{of} \text{acetone}$

${\text{n}}_2\text{=} \text{number} \text{of} \text{moles} \text{of} \text{benzophenone}$

The value of n₁ and n₂ can be calculated as follows:

$$\begin{gathered} n_1=\frac{50.0\text{g}}{58.08{\text{gmol}}^{-1}} \\ =0.861\text{mol} \end{gathered}$$

$$\begin{gathered} n_2=\frac{15.0\text{g}}{182.217\text{gmol}} \\ =0.0823\text{mol} \end{gathered}$$

Calculation of vapor pressure

To calculate the value of vapor pressure, the value mole fraction will be determined first. Mole fraction will be determined by use of the formula:

.$X_1=\frac{n_1}{n_1+n_2}$

Substituting the value of n₁ and n₂ in above formula.

$$\begin{gathered} X_1=\frac{0.861\text{mol}}{0.861\text{mol}+0.0823\text{mol}} \\ =\frac{0.861\text{pad}}{0.9433\text{mol}} \\ =0.9127 \end{gathered}$$

Vapor pressure can be calculated as follow,

$$\begin{gathered} P_1=X_1{P}_1^0 \\ =0.9127\times 0.3270\text{atm} \\ =0.2985\text{atm} \end{gathered}$$