Question

Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.

Answer 1

Hint: Remember that the formula of mole fraction of a solute in a chemical solution is given by:
${{\chi }_{solute}}=\dfrac{number\text{ }of\text{ }moles\text{ }of\text{ }solute}{number\text{ }of\text{ }moles\text{ }of\text{ }solute+\text{ }number\text{ }of\text{ }moles\text{ }of\text{ }solvent}$
Now try to apply this concept to the given question.

Step-by-Step Solution:
Before coming to this question specifically, let us first understand what the term mole fraction and percentage by weight really mean.
Mole fraction is simply the number of moles of a particular substance compared to the total number of moles. Since the number of moles of one substance in a mixture can never be more than the total number of moles, the mole fraction for a substance always ranges between 0 (none) and 1 (all).
The mole fraction of a particular solute in a given solution is given by:
${{\chi }_{solute}}=\dfrac{number\text{ }of\text{ }moles\text{ }of\text{ }solute}{number\text{ }of\text{ }moles\text{ }of\text{ }solute+\text{ }number\text{ }of\text{ }moles\text{ }of\text{ }solvent}$
Now, let us come to the term percentage by mass.
Mass percentage is one way of representing the concentration of an element in a compound or a component in a mixture. Mass percentage is calculated as the mass of a component divided by the total mass of the mixture, multiplied by 100%.
\[Mass\text{ }percent\text{ }=\text{ }\left( \dfrac{Grams\text{ }of\text{ }solute\text{ }}{Grams\text{ }of\text{ }solution} \right)\text{ }\times \text{ 100}\]
Now, let us apply these concepts to the given question.
Let the mass of solution be 100 g
Mass of Benzene = 30 g
Mass of carbon tetrachloride = 100 g - 30 g = 70 g
Number of moles of benzene = Mass / Molar mass
                     = 30 g / 78 g mol⁻¹
                     = 0.385 mol
Number of moles of carbon tetrachloride = Mass / Molar mass
                                                                  = 70 g / 154 g mol⁻¹
                                                                  = 0.455 mol
${{\chi }_{benzene}}=\dfrac{number\text{ }of\text{ }moles\text{ }of\text{ }benzene}{number\text{ }of\text{ }moles\text{ }of\text{ }benzene+\text{ }number\text{ }of\text{ }moles\text{ }of\text{ }CC{{l}_{4}}}=\dfrac{0.385\text{ mol}}{0.840\text{ mol}}$
Therefore, the final mole fraction is 0.458.

Note: Mole fraction is a unit of concentration, defined to be equal to the number of moles of a component divided by the total number of moles of a solution. Because it is a ratio, mole fraction is a unitless expression. The mole fraction of all components of a solution, when added together, will equal 1.