Question

What is the relationship between the mole fraction of a solute$\left( {{X}_{A}} \right)$and its molality (m) if the molar mass of solvent is 100 (g/mol).
a)$\dfrac{({{X}_{A}})}{10(1-{{X}_{A}})}$
b)$\dfrac{({{X}_{A}})}{100(1-{{X}_{A}})}$
c)$\dfrac{10(1-{{X}_{A}})}{({{X}_{A}})}$
d)$\dfrac{10({{X}_{A}})}{(1-{{X}_{A}})}$

Answer 1

Hint: Mole fraction of any solution is defined as the number of moles in one component upon the total number of moles in all components. The components may be the solute or the solvent. The molality of any solution is the number of moles of solute dissolved per kilogram of the solvent used in a solution.

Complete answer:
We have been given to find the relationship between the mole fraction of a solute$\left( {{X}_{A}} \right)$and its molality (m) given that the molar mass of solvent is 100 (g/mol).
We have the expression of mole fraction as number of moles in component A upon the total number of moles in all components A solute and solvent as ${{X}_{A}}=\dfrac{{{n}_{a}}}{{{n}_{a}}+{{n}_{solvent}}}$ , where n is the number of moles.
From the expression of molality we have, $m=\dfrac{{{n}_{A}}\times 1000}{{{n}_{solvent}}\times 100}$ , so, $m=\dfrac{10\times {{n}_{a}}}{{{n}_{solvent}}}$ .
Now, dividing the mole fraction expression with molality we have,
$\dfrac{{{X}_{A}}}{m}=\dfrac{\dfrac{{{n}_{a}}}{{{n}_{a}}+{{n}_{solvent}}}}{\dfrac{10\times {{n}_{a}}}{{{n}_{solvent}}}}$
$\dfrac{{{X}_{A}}}{m}=\dfrac{{{n}_{a}}\times {{n}_{solvent}}}{10{{n}_{a}}\times ({{n}_{a}}+{{n}_{solvent}})}$
Rearranging the above equation and solving for molality we have,
$m=\dfrac{10{{X}_{A}}({{n}_{a}}+{{n}_{solvent}})}{{{n}_{solvent}}}$ , as mole fraction of solvent is$\dfrac{{{n}_{solvent}}}{({{n}_{a}}+{{n}_{solvent}})}={{X}_{solvent}}$ , therefore,
$m=\dfrac{10{{X}_{A}}}{{{X}_{solvent}}}$
As mole fraction of other component (solvent) can be obtained as $1-{{X}_{A}}$, therefore,
$m=\dfrac{10{{X}_{A}}}{1-{{X}_{A}}}$
Hence, the relationship between mole fraction of a solute$\left( {{X}_{A}} \right)$and its molality (m) is$\dfrac{10({{X}_{A}})}{(1-{{X}_{A}})}$.

So, option D is correct.

Note:
The mole fraction of all the components is equal to unity therefore the expression, ${{X}_{solvent}}=1-{{X}_{A}}$. Mole fraction has no unit as it is the fraction. As molality is expressed in kilograms, therefore in the formula of molality, the volume is divided by the factor of 1000, to make it into kilograms.