Chapter 12, Problem 12.92SP

Interpretation Introduction

To determine:

1. Vapor pressure in mmhg on dissolving 5 g benzoic acid in 100 g ethanol(P⁰ = 100.5 mmhg)

EXPLAINATION:

  $\begin{array}{l}{\text{X}}_{\text{bnezoic}}=\frac{\text{5}/\text{122}}{\text{5}/\text{122}+100/46}=0.021\\ {\text{X}}_{etoh}=\frac{100/46}{\text{5}/\text{122}+100/46}=0.979\end{array}$

  $\begin{array}{l}{P}_s={\text{ P}}^0{}_{\text{benzoic}}{\text{X}}_{\text{benzoic}}+{\text{ P}}^0{}_{\text{etoh}}{\text{X}}_{\text{etoh}}\\ {\text{P}}_{\text{s}}=0+\text{ 1}00.\text{5 }\left(0.\text{979}\right)\\ {\text{P}}_{\text{s}}=\text{ 98}.\text{38 mmhg}\end{array}$