Question

A solution is prepared by mixing 50.0 mL of toluene (C₆H₅CH₃, d=0.867 g/cm³) with 125mL of benzene (C₆H₆, d=0.874 g/cm³). Assuming that the volume add on mixing, calculate the mass percent, mole fraction, molality, and molarity of the toluene.

Short answer

Mass percent= 28.4%

Mole fraction = 0.252

Molality= 4.31m

Molarity=2.69 M

Step-by-step solution

Determine the mass percent

$\mathrm{Mass} \mathrm{percent} =\frac{\mathrm{Grams} \mathrm{of} \mathrm{solute}}{\mathrm{Grams} \mathrm{of} \mathrm{solution}}\times 100$

$\mathrm{Grams} \mathrm{of} \mathrm{toluene}=\mathrm{Density} \mathrm{of} \mathrm{toluene}\times \mathrm{Volume} \mathrm{of} \mathrm{toluene}$

where,

Grams of toluene=43.35

Gramsof solutions=43.35+109.25=152.6

Therefore, mass percent

$$\begin{gathered} =\frac{43.35}{152.6}\times 100 \\ =28.4\% \end{gathered}$$

Determine the mole fraction

$$\begin{gathered} \mathrm{mole} \mathrm{fraction} \mathrm{of} \mathrm{a} \mathrm{component} {\mathrm{X}}_{\mathrm{a}}=\frac{{\mathrm{n}}_{\mathrm{a}}}{{\mathrm{n}}_{\mathrm{a}}+{\mathrm{n}}_{\mathrm{b}}} \\ {\mathrm{n}}_{\mathrm{a}}=\mathrm{number}\mathrm{of}\mathrm{moles}\mathrm{of}\mathrm{component} \\ {\mathrm{n}}_{\mathrm{a}}+{\mathrm{n}}_{\mathrm{b}}= \mathrm{number}\mathrm{of}\mathrm{moles}\mathrm{of}\mathrm{mixture} \end{gathered}$$

Number of moles of toluene

$\frac{43.35}{92.14}=0.470 \mathrm{mol}$

Mole fraction of toluene will be

$\begin{array}{rcl}{\mathrm{X}}_{\mathrm{toluene}}& =& \frac{\mathrm{no} \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{toluene}}{\mathrm{no} \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{toluene}+\mathrm{no} \mathrm{of} \mathrm{moles} \mathrm{of} \mathrm{benzene}}\\ & =& \frac{0.4705}{0.4705+1.399}\\ & =& 0.252\end{array}$

Determine the molality

$\begin{array}{rcl}\mathrm{Molality} \mathrm{of} \mathrm{toluene}& =& \frac{\mathrm{No} \mathrm{of} \mathrm{mols} \mathrm{of} \mathrm{toluene}}{\mathrm{Amount} \mathrm{of} \mathrm{toluene} (\mathrm{in} \mathrm{Kg})}\\ & =& \frac{0.4705 \mathrm{mol} }{109.25 \mathrm{g}\times \frac{1 \mathrm{Kg}}{1000 \mathrm{g}}}\\ & =& 4.31 \mathrm{m}\end{array}$

Determine the molarity

$\begin{array}{rcl}\mathrm{Molarity} \mathrm{of} \mathrm{toluene}& =& \frac{\mathrm{No} \mathrm{of} \mathrm{mols} \mathrm{of} \mathrm{toluene}}{\mathrm{Volume} \mathrm{of} \mathrm{Solution} \mathrm{in} \mathrm{L}}\\ & =& \frac{0.4705 \mathrm{mol} }{\left(50+125\right)\mathrm{mL}\times \frac{1 \mathrm{L}}{1000 \mathrm{mL}}}\\ & =& 2.69 \mathrm{M}\end{array}$