Question

The vapor pressure of diethyl ether (molar mass, $\mathbf{74}\mathbf{.}\mathbf{12}\mathbf{}{\mathbf{\text{gmol}}}^{\mathbf{-}\mathbf{1}}$at${\mathbf{30}}^{\mathbf{0}}\mathit{C}$ is 0.8517 atm. Suppose 1.800 gof maleic acid,${\mathbf{\text{C}}}_{\mathbf{4}}{\mathbf{\text{H}}}_{\mathbf{4}}{\mathbf{\text{O}}}_{\mathbf{4}}\mathbf{,}$is dissolved in 100.0 gof diethyl ether at$\mathbf{30}\mathbf{.}{\mathbf{0}}^{\mathbf{0}}\mathit{C}$Calculate the vapor pressure of diethyl ether above the resulting solution.

Short answer

The vapor pressure of diethyl ether above the resulting solution is $\text{0.8420 atm}$.

Step-by-step solution

Given data

Vapor pressure of diethyl ether is 0.8517 atm.

The amount of maleic acid is 1.8 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

Formula of vapor pressure

The reaction between vapor pressure and mole fraction can be shown as,

$P_1=X_1{P^0}_1$ ……. (i)

where, P₁i s the vapor pressure of the solution, X₁is the mole fraction of the solute, P⁰₁is the vapor pressure of the pure solvent.

Calculation of n1 and n2

${\text{n}}_1\text{=} \text{number} \text{of} \text{moles} \text{of} \text{diethyl ether}$

${\text{n}}_2\text{=} \text{number} \text{of} \text{moles} \text{of} \text{maleic} \text{acid}$

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

$$\begin{gathered} n_1=\frac{100.0\text{g}}{74.12{\text{gmol}}^{-1}} \\ =1.349\text{mol} \end{gathered}$$

$$\begin{gathered} n_2=\frac{1.800\text{g}}{116.07{\text{gmol}}^{-1}} \\ =0.0155\text{mol} \end{gathered}$$

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 n₁and n₂ in above formula.

$$\begin{gathered} X_1=\frac{1.349\text{mol}}{1.349\text{mol}+0.0155\text{mol}} \\ =\frac{1.349\text{mol}}{1.3645\text{mol}} \\ =0.9886 \end{gathered}$$

Vapor pressure can be calculated as follows:

$$\begin{gathered} P_1=X_1{P^0}_1=0.9886\times 0.8517\text{atm} \\ =0.8420\text{atm} \end{gathered}$$