Chapter 13, Problem 13.108SP

Interpretation Introduction

Interpretation:

Vapor pressure of pure A and B solutions at $\text{84°C}$ have to be calculated.

Concept introduction:

Raoult’s law states that in an ideal mixture of liquid solution, partial pressure of every component is equal to its mole fraction multiplied into vapour pressure of its pure components.

${\text{P}}_{}{\text{= χ}}_{}{\text{P}}^{\text{°}}$

Where,

$\begin{array}{l}\text{P- Partial pressure of each component }\\ \text{P°-Partial pressure of its pure components }\\ \text{χ- Mole fraction of the components }\end{array}$

Answer to Problem 13.108SP

Vapor pressure of pure A solution = $\text{1}{\text{.9×10}}^{\text{2}}\text{mmHg}$

Vapor pressure of pure B solution = $\text{4}{\text{.0×10}}^{\text{2}}\text{mmHg}$

Record the given data

Vapor pressure of the solution = $\text{331mmHg}$

Amount of solution A =$\text{1}\text{.2 moles}$

Amount of solution B =$\text{2}\text{.3 moles}$

Increased vapor pressure of the solution = $\text{347 mmHg}$

Explanation of Solution

Given data are recorded as shown.

To calculate mole fraction of total solution

$\begin{array}{l}{\text{χ}}_{\text{A}}\text{=}\frac{\text{1}\text{.2mol}}{\text{1}\text{.2mol+2}\text{.3mol}}\text{= 0}\text{.3429}\\ \\ {\text{χ}}_{\text{B}}\text{=1- 0}\text{.3429 = 0}\text{.6571}\\ \\ {\text{P}}_{\text{total}}{\text{= P}}_{\text{A}}{\text{+P}}_{\text{B}}{\text{= χ}}_{\text{A}}{\text{P}}_{\text{A}}^{\text{°}}{\text{+ χ}}_{\text{B}}{\text{P}}_{\text{B}}^{\text{°}}\end{array}$

Substituting in ${\text{P}}_{\text{total}}$ and the mole fractions,

$\text{331mmHg = 0}{\text{.3429P}}_{\text{A}}^{\text{°}}\text{+0}{\text{.6571P}}_{\text{B}}^{\text{°}}$

${\text{P}}_{\text{A}}^{\text{°}}\text{=}\frac{\text{331mmHg - 0}{\text{.6571P}}_{\text{B}}^{\text{°}}}{\text{0}\text{.3429}}\text{ = 965}\text{.3mmHg -1}{\text{.916P}}_{\text{B}}^{\text{°}}$ $\mathrm{......}(1)$

By plugging in the values of mole fraction of solution A and B, mole fraction of total solution has calculated.

To calculate mole fraction of total solution after additional moles of B

 $\begin{array}{l}{\text{χ}}_{\text{A}}\text{=}\frac{\text{1}\text{.2mol}}{\text{1}\text{.2mol+3}\text{.3mol}}\text{= 0}\text{.2667}\\ \\ {\text{χ}}_{\text{B}}\text{=1- 0}\text{.2667 = 0}\text{.7333}\\ \\ {\text{P}}_{\text{total}}{\text{= P}}_{\text{A}}{\text{+P}}_{\text{B}}{\text{= χ}}_{\text{A}}{\text{P}}_{\text{A}}^{\text{°}}{\text{+ χ}}_{\text{B}}{\text{P}}_{\text{B}}^{\text{°}}\end{array}$

Substituting in ${\text{P}}_{\text{total}}$ and the mole fractions,

$\text{347 mmHg = 0}\text{.2667(965}\text{.3 mmHg-1}{\text{.916P}}_{\text{B}}^{\text{°}}\text{) + 0}{\text{.7333P}}_{\text{B}}^{\text{°}}$$\mathrm{......}(2)$

By plugging in the values of mole fraction of solution A and mole fraction of solution after addition of solution B, after additional moles of B has calculated.

To calculate vapor pressure of pure solution of A and B

Substituting equation (1) into (2) we get,

$\text{347mmHg = 0}\text{.2667(965}\text{.3mmHg -1}{\text{.916P}}_{\text{B}}^{\text{°}}\text{) + 0}{\text{.7333P}}_{\text{B}}^{\text{°}}$

$\begin{array}{l}\text{0}{\text{.2223P}}_{\text{B}}^{\text{°}}\text{= 89}\text{.55mmHg}\\ {\text{P}}_{\text{B}}^{\text{°}}\text{= 402}\text{.8mmHg = 4}{\text{.0×10}}^{\text{2}}\text{mmHg}\end{array}$

Substituting the value of

${\text{P}}_{\text{A}}^{\text{°}}\text{= 965}\text{.3mmHg -1}\text{.916(402}\text{.8mmHg) =193}\text{.5mmHg =1}{\text{.9×10}}^{\text{2}}\text{mmHg}$

By substituting the equation (1) into equation (2), the vapor pressure of pure solution B has calculated and by subtracting this value into ${\text{P}}_{\text{A}}^{\text{°}}$, the vapor pressure of pure solution B has calculated.

Conclusion

Vapor pressure of pure solution A was calculated as $\text{1}{\text{.9×10}}^{\text{2}}\text{mmHg}\text{.}$

Vapor pressure of pure solution B was calculated as $\text{4}{\text{.0×10}}^{\text{2}}\text{mmHg}\text{.}$