Question

In a litre flask at a certain temperature, there are 2 gm of ${H_2}$ and 8 gm of ${O_2}$. The mole fraction of ${O_2}$ in the given mixture is:
A.$0.2$
B.$0.4$
C.$0.3$
D.$0.1$

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,
the mole fraction using the formula,
${\chi _A} = \dfrac{{{\text{Moles of A }}\left( {{\text{in mol}}} \right)}}{{{\text{Moles of A}}\left( {{\text{in mol}}} \right) + {\text{Moles of B}}\left( {{\text{in mol}}} \right) + {\text{Moles of C}}\left( {{\text{in mol}}} \right)}}$
${\chi _A} = \dfrac{{{\text{Moles of A }}\left( {{\text{in mol}}} \right)}}{{{\text{Total mass}}}}$

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,
${\chi _A} = \dfrac{{{\text{Moles of A }}\left( {{\text{in mol}}} \right)}}{{{\text{Moles of A}}\left( {{\text{in mol}}} \right) + {\text{Moles of B}}\left( {{\text{in mol}}} \right)}}$
${\chi _A} = \dfrac{{{\text{Moles of A }}\left( {{\text{in mol}}} \right)}}{{{\text{Total mass}}}}$
Now, coming back to the question, we have to calculate the mole fraction of oxygen. We can calculate the mole fraction of oxygen using the moles of oxygen and moles of hydrogen.
Given data contains,
Mass of hydrogen is \[2gm\].
Mass of oxygen is \[8gm\].
Let us now calculate the moles of oxygen and hydrogen using their molecular masses.
The molecular mass of oxygen is \[32g/mol\].
The molecular mass of hydrogen is \[1g/mol\].
The moles of oxygen is calculated as,
Moles of oxygen $ = \dfrac{{{\text{Grams}}}}{{{\text{Molecular mass}}}}$
Moles of oxygen=$\dfrac{{8g}}{{32g/mol}}$
Moles of oxygen=$0.25mol$
The moles of oxygen is $0.25mol$
The moles of hydrogen is calculated as,
Moles of hydrogen $ = \dfrac{{{\text{Grams}}}}{{{\text{Molecular mass}}}}$
Moles of hydrogen=$\dfrac{{2g}}{{1g/mol}}$
Moles of hydrogen=$1mol$
The moles of hydrogen is $1mol$.
From the moles of hydrogen and oxygen, we can calculate the mole fraction of oxygen as, ${\chi _{\left( {oxygen} \right)}} = \dfrac{{{\text{Moles of oxygen }}\left( {{\text{in mol}}} \right)}}{{{\text{Total mass}}}}$
${\chi _{oxygen}} = \dfrac{{0.25mol}}{{0.25mol + 1mol}}$
${\chi _{oxygen}} = 0.2mol$
The mole fraction of oxygen is $0.2mol$.
Therefore, Option (A) is correct.


Note:We can calculate mole fraction from mass percent, molarity, 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,
${P_{solv}} = {\chi _{solv}}{P^o}_{solv}$
The vapor pressure of the solvent $\left( {{P_{solv}}} \right)$ above a dilute solution is equal to the mole fraction of the solvent $({\chi _{solv}})$ times the vapor pressure of the pure solvent $\left( {{P_{solv}}^o} \right)$.