Question

Write the formula of mole fraction.

Answer 1

Hint – The method of mole fraction is utilized when the solution is constituted by mixing two or more components. Mole fraction is equivalent to the ratio of moles of a specific component to the number of moles of solution. Moles of solution is equivalent to the moles of solute and moles of solvent.

Step-by-step answer:
The mole fraction is designated by the symbol “$\chi$” which is referred to as chi. The mole fraction is the ratio of the number of moles of one component to the total number of moles of the solution (i.e., all the components).

Let’s assume that solute B having mass “${w_{\text{B}}}$” and molar mass “${M_{\text{B}}}$” dissolves in solvent A having mass “${w_{\text{A}}}$” and molar mass “${M_{\text{A}}}$”.

The number of moles of solute $\left( {{n_{\text{B}}}} \right)$ and solvent $\left( {{n_{\text{A}}}} \right)$ can be obtained by using the following formula.
${\text{Number}}\;{\text{of}}\;{\text{moles}} = \dfrac{{{\text{Mass}}}}{{{\text{Molar}}\;{\text{mass}}}} \cdot \cdot \cdot \cdot \cdot \cdot \left( {\text{I}} \right)$

For solute B, mass is ${w_{\text{B}}}$ and molar mass is ${M_{\text{B}}}$. Thus on substituting these variables in equation (I), we get
${n_{\text{B}}} = \dfrac{{{w_{\text{B}}}}}{{{M_{\text{B}}}}} \cdot \cdot \cdot \cdot \cdot \cdot \left( {{\text{II}}} \right)$

For solvent A, mass is ${w_{\text{A}}}$ and molar mass is ${M_{\text{A}}}$. Thus on substituting these variables in equation (I), we get
${n_{\text{A}}} = \dfrac{{{w_{\text{A}}}}}{{{M_{\text{A}}}}} \cdot \cdot \cdot \cdot \cdot \cdot \left( {{\text{III}}} \right)$

The mole fraction of a solvent is obtained by dividing moles of solvent to the moles of solution (all components). Thus the formula of mole fraction of solvent $\left( {{\chi _{\text{A}}}} \right)$ is obtained as:
${\chi _{\text{A}}} = \dfrac{{{n_{\text{A}}}}}{{{n_{\text{A}}} + {n_{\text{B}}}}} \cdot \cdot \cdot \cdot \cdot \cdot \left( {{\text{IV}}} \right)$

By substituting equations (II) and (III) in equation (IV), the formula of mole fraction of solvent $\left( {{\chi _{\text{A}}}} \right)$ can also be written as follows:
${\chi _{\text{A}}} = \dfrac{{\dfrac{{{w_{\text{A}}}}}{{{M_{\text{A}}}}}}}{{\dfrac{{{w_{\text{A}}}}}{{{M_{\text{A}}}}} + \dfrac{{{w_{\text{B}}}}}{{{M_{\text{B}}}}}}}$

Similarly, the mole fraction of a solute is obtained by dividing moles of solute to the moles of solution (all components). Thus the formula of mole fraction of solute $\left( {{\chi _{\text{B}}}} \right)$ can be obtained as:
${\chi _{\text{B}}} = \dfrac{{{n_{\text{B}}}}}{{{n_{\text{A}}} + {n_{\text{B}}}}} \cdot \cdot \cdot \cdot \cdot \cdot \left( {\text{V}} \right)$

By substituting equations (II) and (III) in equation (V), the formula of mole fraction of solute $\left( {{\chi _{\text{B}}}} \right)$ can also be written as follows:
${\chi _{\text{B}}} = \dfrac{{\dfrac{{{w_{\text{B}}}}}{{{M_{\text{B}}}}}}}{{\dfrac{{{w_{\text{A}}}}}{{{M_{\text{A}}}}} + \dfrac{{{w_{\text{B}}}}}{{{M_{\text{B}}}}}}}$

Note: For a binary solution, the sum of mole fractions of a solute and solvent (i.e., solution) is equal to 1. Thus it can be said that:

{\chi _{\text{A}}} + {\chi _{\text{B}}} = 1 \cr {\chi _{\text{B}}} = 1 - {\chi _{\text{A}}} \cr} $$