Question

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

Answer 1

Mole fraction can be defined as the number of moles of a particular component (solute or solvent) in a solution divided by the total number of moles in the given solution. The number of moles can be calculated by dividing the given mass of the compound with the molar mass.

Complete step by step answer:
Mole fraction is a term of concentration that is used to relatively measure the concentrations of solute and solvent in a mixture solution.
$Mole\,fraction\,of\,the\,solute\,\, = \,\,\dfrac{{No.\,of\,moles\,of\,solute}}{{No.\,of\,moles\,of\,solute\, + \,No.\,of\,moles\,of\,solvent\,}}$
If the no of moles of solute is denoted by ${n_A}$ and no of moles of solvent by ${n_B}$, then the mole fraction of solute and solvent can be denoted as ${X_A}$ or ${X_B}$ and we can represent them as
${X_{A\,}}\, = \,\,\dfrac{{{n_A}}}{{{n_A}\, + \,{n_B}}}$
Where ${X_A}$ is the Mole fraction of solute or component A.
${X_{B\,}}\, = \,\dfrac{{{n_B}}}{{{n_A}\, + \,{n_B}}}$
Where ${X_B}$ is the mole fraction of solvent or component B.
We are asked to calculate the mole fraction of benzene which is an aromatic compound having a molecular formula of ${C_6}{H_6}$ .So first we will calculate the no of moles of both the component and then we will apply the above mentioned formula to find mole fraction of benzene.
The molecular mass of benzene will be
$(12 \times 6)\, + \,(1 \times 6)\, = \,78\,g/mol.$
The given mass of benzene is $30\,g$ as it is present $30\%$ by mass which means $30\,g$ of substance is present in total $100g$ of mixture so the given mass of solvent, which is $CC{l_4}$ will be
$100\, - \,30\, = \,70\,g$
And the molecular mass of $CC{l_4}$ will be
$(1 \times 12)\, + \,(35.4 \times 4)\,\, \approx \,\,154\,g/mol.$
Now the no of moles of benzene will be
$= \,\dfrac{{given\,mass\,of\,benzene}}{{molar\,mass\,of\,benzene}}$
$\,\,\,{n_A}\, = \,\,\dfrac{{30}}{{78}}\,\, = \,\,\,0.385$
And the no of moles of $CC{l_4}$ will be
${n_B}\,\, = \,\dfrac{{70}}{{154}}\,\, = \,\,0.45$
And now the mole fraction of benzene will be
${X_A}\, = \,\dfrac{{No.\,of\,moles\,of\,benzene\,}}{{No.\,of\,moles\,of\,benzene\, + \,No.\,of\,moles\,of\,CC{l_4}}}$
${X_A}\, = \,\dfrac{{0.385}}{{0.385 + 0.45}} \\\ \\\$
${X_A}\, = \,0.461$

Note:
The molecular mass of a given compound is the total mass of one mole of that compound and one mole of a substance means the value of $6.022 \times \,{10^{23}}$ particles of that compound. Other terms that are used to determine the concentration are Molarity and Molality etc.