Question

Calculate the mole fraction of benzene in solution containing $30\%$ by mass in $CC{l_4}$?

Answer 1

Mole fraction of a component present in mixture is equal to the ratio of the moles of pure component to the total moles present in the mixture.

Complete step by step answer:
It is given that the component is benzene which is present in $CC{l_4}$ solution. The mass percentage of benzene is $30\%$. The mass percentage indicates the amount of a substance in grams present in $100mL$ of solution.
Thus the amount of benzene is $30g$. Let the total weight of the mixture is $100g$.
Therefore the weight of $CC{l_4}$ is ($100 - 30$) = $70g$.
$w{}_{benzene}$ = $30g$ and ${w_{CC{l_4}}}$= $70g$.
Mole fraction is the ratio of mole of pure component and mole of total mixture of components.
${\chi _A} = \dfrac{{{n_A}}}{{{n_T}}} = \dfrac{{{n_A}}}{{{n_A} + {n_B} + {n_C} + ......}}$
Where ${\chi _A}$is the referred as mole fraction of the component $A$,
${n_A}$is the mole of component $A$, ${n_T}$is the mole of the total mixture of components.
Thus, in order to determine the mole fraction of benzene, the moles of benzene and $CC{l_4}$ have to be evaluated first.
The mole of the component can be obtained by the ratio of the weight of the component and the molar mass of that component.
The molar mass of benzene is $78g/mol$ and the molar mass of $CC{l_4}$ is $154g/mol$.
Thus the mole of benzene = $\dfrac{{30}}{{78}} = 0.385mol$
The mole of $CC{l_4}$ = $\dfrac{{70}}{{154}} = 0.455mol.$
The mole fraction of benzene = $\dfrac{{{n_{benzene}}}}{{{n_{benzene}} + {n_{CC{l_4}}}}}$
${\chi _{benzene}} = \dfrac{{0.385}}{{0.385 + 0.455}}$
${\chi _{benzene}} = \dfrac{{0.385}}{{0.84}}$
${\chi _{benzene}} = 0.458$
Hence, the mole fraction of benzene present in solution which contains $30\%$ by mass in $CC{l_4}$is 0.458.

Note: Please take care of the unit of mole and mole fraction. Mole fraction is unitless. The total component can be two, three, and four and so on. But the mole of each component is required to determine the mole fraction.