Chapter 4, Problem 4D.5E

Interpretation Introduction

Interpretation:

The mole fractions of the solute and solvent for a solute of molality $0.100molk{g}^{-1}$ has to be determined and so the change in the chemical potential of the solvent caused by the solute has to be calculated.

Concept introduction:

Chemical potential:

The chemical potential tell us how the Gibbs Free energy will vary with the number of moles $n_i$ of component i holding temperature, pressure, and the number of moles of all other components constant. For a pure substance, the chemical potential is equal to its molar Gibbs Free energy.

${\text{μ}}_{\text{i}}\text{=}{\left(\frac{\partial \text{G}}{\partial {\text{n}}_{\text{i}}}\right)}_{{\text{T,P,n}}_{\text{i,j}}}\boxed{\begin{array}{l}\text{where,}\\ {\text{μ}}_{\text{i}}\text{Chemical}\text{potential}\\ \text{G}\text{Gibbs}\text{free}\text{energy}\\ \text{T}\text{Temperature}\\ \text{P}\text{Pressure}\end{array}}$

Molality (m): Molality is the number of moles of solute present in one kilogram of solvent.

$\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1kg}\text{of}\text{solvent}}$

A mole fraction of a molecule in a mixture is the ratio of number of moles of particular molecule to the sum of number of moles of all molecules in the mixture. Equation for mole fraction of a molecule in a mixture of three molecules (A, B and C) is,

  $\text{molecule}\text{fraction}\text{of}\text{A,}\text{(}{\text{χ}}_{\text{A}}\text{) }\text{=}\frac{\text{numbers}\text{of moles of}\text{molecule A}{\text{(n}}_{\text{A}}\text{)}}{\text{total number of moles}{\text{(n}}_{\text{A}}{\text{+n}}_{\text{B}}{\text{+n}}_{\text{C}}\text{) }}$

The equation which relates the molality with the mole fraction of solute is,

$\begin{array}{l}Molality\left(m\right)=\frac{x_B}{1-x_B}\times \frac{1000}{\left(molarmassofsolvent\right)}\boxed{\begin{array}{l}Where,\\ x_B:molefractionofsolute\end{array}}\\ \end{array}$