Question

Calculate the mole fraction of Benzene $\left( {{C_6}{H_6}} \right)$ in a solution containing 30% mass in $HCl$$\left( {Cl = 35.5} \right)$

Answer 1

Hint: We can calculate the mole fraction of a substance in a solution by calculating the moles of each solute dividing by total number of moles of the solution. We can calculate the moles of the solute by converting the grams of the solute to moles of the solute using molar mass. The total moles of solution are calculated by adding the moles of each solute/solvent.

Formula used: We can calculate the mole fraction using the formula,
${\chi _A} = \dfrac{{{\text{Moles of A}}\left( {{\text{in mol}}} \right)}}{{{\text{Moles of A}}\left( {{\text{in mol}}} \right) + {\text{Moles of B}}\left( {{\text{in mol}}} \right) + {\text{Moles of C}}\left( {{\text{in mol}}} \right)}}$
${\chi _A} = \dfrac{{{\text{Moles of A}}\left( {{\text{in mol}}} \right)}}{{{\text{Total number of moles of components}}\left( {{\text{in mol}}} \right)}}$

Complete step by step answer:
We can define Mole fraction as the ratio of moles of one substance in a mixture to the total number of moles of all substances. For a solution of two substances A and B, we can calculate the mole fraction using the formula,
${\chi _A} = \dfrac{{{\text{Moles of A}}\left( {{\text{in mol}}} \right)}}{{{\text{Moles of A}}\left( {{\text{in mol}}} \right) + {\text{Moles of B}}\left( {{\text{in mol}}} \right)}}$
${\chi _A} = \dfrac{{{\text{Moles of A}}\left( {{\text{in mol}}} \right)}}{{{\text{Total number of moles of components}}\left( {{\text{in mol}}} \right)}}$
${\chi _A} = \dfrac{{{\text{Moles of B}}\left( {{\text{in mol}}} \right)}}{{{\text{Moles of A}}\left( {{\text{in mol}}} \right) + {\text{Moles of B}}\left( {{\text{in mol}}} \right)}}$
${\chi _A} = \dfrac{{{\text{Moles of B}}\left( {{\text{in mol}}} \right)}}{{{\text{Total number of moles of components}}\left( {{\text{in mol}}} \right)}}$
Now, coming back to the question, we have to calculate the mole fraction of benzene. We can calculate the mole fraction of benzene using the moles of benzene and moles of hydrogen chloride.
Let us consider the mass of the solution to be $100g$ .
Mass of benzene is $30g\left( {30\% = 30g} \right)$ .
Mass of hydrogen chloride is calculated using the difference between the mass of the solution and mass of benzene.
We can calculate the mass of hydrogen chloride as follows,
Mass of hydrogen chloride = Mass of solution-Mass of benzene
Mass of hydrogen chloride = $100g - 30g$
Mass of hydrogen chloride = ${\text{70g}}$
The mass of hydrogen chloride is ${\text{70g}}$.
We can now calculate the moles of benzene as follows,
Molar mass of benzene = $78g/mol$.
Moles of benzene = $\dfrac{{{\text{Mass of benzene}}\left( {{\text{in grams}}} \right)}}{{{\text{Molar mass}}}}$
Moles of benzene = $\dfrac{{{\text{30g}}}}{{{\text{78g/mol}}}}$
Moles of benzene = ${\text{0}}{\text{.3846mol}}$
The moles of benzene is ${\text{0}}{\text{.3846mol}}$.
We can now calculate the moles of hydrogen chloride as follows,
Molar mass of hydrogen chloride = $35.5g/mol$.
Moles of hydrogen chloride = $\dfrac{{{\text{Mass of hydrogen chloride}}\left( {{\text{in grams}}} \right)}}{{{\text{Molar mass}}}}$
Moles of hydrogen chloride = $\dfrac{{{\text{70g}}}}{{{\text{35}}{\text{.5g/mol}}}}$
Moles of hydrogen chloride = ${\text{1}}{\text{.971mol}}$
The moles of hydrogen chloride is ${\text{1}}{\text{.971mol}}$.
We can calculate the mole fraction of benzene using the moles of hydrogen chloride.
The mole fraction of benzene is calculated as,
Mole fraction of benzene = $\dfrac{{{\text{Moles of benzene}}\left( {{\text{in mol}}} \right)}}{{{\text{Moles of benzene}}\left( {{\text{in mol}}} \right) + {\text{Moles of HCl}}\left( {{\text{in mol}}} \right)}}$
Mole fraction of benzene = $\dfrac{{{\text{0}}{\text{.385mol}}}}{{{\text{0}}{\text{.385mol + 1}}{\text{.971mol}}}}$
Mole fraction of benzene = ${\text{0}}{\text{.1634}}$
The mole fraction of benzene is ${\text{0}}{\text{.1634}}$.

Note: We can calculate mole fraction from mass percent, molarity, and molality. Some of the properties of mole fraction are,
It does not depend on temperature.
A mixture of mole fractions could be prepared by weighting the masses of the substances.
We can also calculate the mole fraction from Raoult’s law. The formula we use to calculate the mole fraction from Raoult’s law is,
${{\text{P}}_{{\text{solv}}}}{\text{ = }}{{{\chi }}_{{\text{solv}}}}{{\text{P}}^{\text{o}}}_{{\text{solv}}}$
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 ${\chi _{solv}}$ times the vapor pressure of the pure solvent $\left( {{{\text{P}}_{{\text{solv}}}}^{\text{o}}} \right)$.