Chapter 5, Problem 5C.4E

(a)

Interpretation Introduction

Interpretation:

The vapor pressure of the solvent in an aqueous solution of ${\text{40}}^{\text{o}}\text{C}$ in which the mole fraction of glucose is $\text{0}\text{.0316}$ has to be calculated.

Concept Introduction:

The equilibrium between a liquid and its vapor produces a characteristic vapor pressure for each substance that depends on the temperature.  The lowering of the vapor pressure is caused by a lesser ability of the solvent to evaporate, so equilibrium is reached with a smaller concentration of the solvent in the gas phase.  The vapor pressure of a solution is expressed using Raoult’s law:

  ${\text{P}}_{\text{solv}}{\text{=χ}}_{\text{solv}}{\text{P}}^{\text{o}}{}_{\text{solv}}$

The vapor pressure of the solvent $\left({\text{P}}_{\text{solv}}\right)$ above a dilute solution is equal to the mole fraction of the solvent ${\text{(χ}}_{\text{solv}}\text{)}$ times the vapor pressure of the pure solvent $\left({\text{P}}^{\text{o}}{}_{\text{solv}}\right).$

Mole fraction:  Mole fraction of a substance in a solution is the number of moles of that substance divided by the total number of moles of all substances present.  The formula is,

  $\begin{array}{l}{\text{χ}}_{\text{A}}\text{=}\frac{\text{Moles}\text{of}\text{A (in}\text{mol)}}{\text{Moles}\text{of}\text{A (in mol) +Moles}\text{of}\text{B (in mol) +Moles}\text{of}\text{C (in mol) +}\mathrm{...}}\\ \\ {\text{χ}}_{\text{A}}\text{=}\frac{\text{Moles}\text{of}\text{A (in}\text{mol)}}{\text{Total}\text{number}\text{of}\text{moles}\text{of}\text{components}\text{(in}\text{mol)}}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The vapor pressure of the solvent in an aqueous solution of ${\text{23}}^{\text{o}}\text{C}$ in which the concentration of urea is $\text{0}\text{.0}\text{.240}\text{M}$ has to be calculated.

Concept Introduction:

Refer to part (a).