Question

Iodine is dissolved in ${{C}_{6}}{{H}_{6}}$​ solution. The mole fraction of ${{I}_{2}}$ ​ in solution is 0.25. Find out the molality of the solution.

Answer 1

. Let $\chi$ be the mole fraction of the solute and $m$ be the molality of the solution, then the relationship between $m$ and $\chi$ is given by:
$m=\dfrac{\chi \times 1000}{(1-\chi )\times {{M}_{solvent}}}$
Where, ${{M}_{solvent}}$ is the mass of the solvent given in the question.
Use this relation to try and work out the answer to this question.

Complete step by step answer:
Let us first understand the concepts of mole fraction and molality individually before moving onto the derivation of the relationship between the two.
The mole fraction of a substance is the ratio of moles of a particular component to the total number of moles of a solution.

If a substance $A$ dissolves in substance $B$ and their number of moles are ${{n}_{A}}$ and ${{n}_{B}}$ respectively; then the mole fractions of $A$ and $B$ are given as:
Mole fraction of $A$ = ${{\chi }_{A}}$
= $\dfrac{\text{no}\text{. of moles of }A}{\text{no}\text{. of moles of }A\text{ + no}\text{. of moles of }B}$
= $\dfrac{{{n}_{A}}}{{{n}_{A}}+{{n}_{B}}}$
Mole fraction of $B$ = ${{\chi }_{B}}$
= $\dfrac{\text{no}\text{. of moles of }B}{\text{no}\text{. of moles of }A\text{ + no}\text{. of moles of }B}$
=$\dfrac{{{n}_{B}}}{{{n}_{A}}+{{n}_{B}}}$
Remember that,
${{\chi }_{A}}+{{\chi }_{B}}=1$

Let us now move onto the definition of molality.
Molality is defined as the number of moles of solute dissolved in $1kg(1000g)$ of solvent. The units of molality are given as $mol/kg$ or $m$ , which represent the number of moles of solute per kilogram of solvent.
The primary advantage of using molality to specify concentration over volume-centric measures of concentration like molarity is that unlike its volume, the mass of the solvent remains constant and does not change with changes in temperature or pressure.
Therefore, molality remains constant under changing environmental conditions.

With these ideas in mind, let us now try to solve the given question.
Mole fraction = 0.25
$\therefore$ In 100 moles 25 moles of ${{I}_{2}}$​ & 75 moles are ${{C}_{6}}{{H}_{6}}$
$\therefore$ Mass of benzene= $75\times 78 = 5850g$ (Molar mass of benzene = 78)
$\therefore$ For 5850g of solvent we have 25 moles of ${{I}_{2}}$​.
$\therefore$ For 1000g of solvent we have,
$Molality = \dfrac{1000\times 25}{5850} = 4.27\text{ }mol/kg$
Hence molality of the solution = $4.27mol/kg$

Note: Although their spellings are similar, molarity and molality cannot be interchanged. Molarity is a measurement of the moles in the total volume of the solution, whereas molality is a measurement of the moles in relationship to the mass of the solvent.