A solution is prepared by mixing \(50.0 \mathrm{~mL}\) toluene \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3},\right.\), \(\left.d=0.867 \mathrm{~g} / \mathrm{cm}^{3}\right)\) with \(125 \mathrm{~mL}\) benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}, d=0.874 \mathrm{~g} / \mathrm{cm}^{3}\right)\) Assuming that the volumes add on mixing, calculate the mass percent, mole fraction, molality, and molarity of the toluene.

Using the given volumes and densities, the mass of each component can be calculated as follows: Mass of toluene = volume \(\times\) density Mass of toluene = \(\left(50.0 \mathrm{~mL}\right) \left(0.867 \mathrm{~g} / \mathrm{cm}^{3}\right) = 43.35 \mathrm{~g}\) (note that 1 mL = 1 cm³) Mass of benzene = volume \(\times\) density Mass of benzene = \(\left(125 \mathrm{~mL}\right) \left(0.874 \mathrm{~g} / \mathrm{cm}^{3}\right) = 109.25 \mathrm{~g}\)

The molar mass of each component is calculated by summing the atomic masses of the elements in the compounds' formulas: Molar mass of toluene (C₁₆H₀₆) = 12.01 × 7 + 1.01 × 8 = 84.14 g/mol Molar mass of benzene (C₆H₆) = 12.01 × 6 + 1.01 × 6 = 78.12 g/mol

Divide the mass of each component by its molar mass to find the number of moles: Moles of toluene = \(\frac{43.35 \mathrm{~g}}{84.14 \mathrm{~g/mol}} = 0.515 \mathrm{~mol}\) Moles of benzene = \(\frac{109.25 \mathrm{~g}}{78.12 \mathrm{~g/mol}} = 1.398 \mathrm{~mol}\)

Using the calculated masses and moles, we can now find the desired quantities: Mass percent of toluene = \(\frac{\text{mass of toluene}}{\text{total mass of solution}} \times 100\) Mass percent of toluene = \(\frac{43.35 \mathrm{~g}}{43.35 \mathrm{~g} + 109.25 \mathrm{~g}} \times 100 = 28.4 \%\) Mole fraction of toluene (X_toluene) = \(\frac{\text{moles of toluene}}{\text{total moles}}\) Mole fraction of toluene = \(\frac{0.515 \mathrm{~mol}}{0.515 \mathrm{~mol} + 1.398 \mathrm{~mol}} = 0.269\) Molality of toluene = \(\frac{\text{moles of toluene}}{\text{mass of benzene (in kg)}}\) Molality of toluene = \(\frac{0.515 \mathrm{~mol}}{0.10925 \mathrm{~kg}} = 4.71 \mathrm{~mol/kg}\) Molarity of toluene = \(\frac{\text{moles of toluene}}{\text{total volume of solution (in L)}}\) Molarity of toluene = \(\frac{0.515 \mathrm{~mol}}{0.175 \mathrm{~L}} = 2.94 \mathrm{~M}\) From the calculations above, we find that the mass percent of toluene in the solution is 28.4%, the mole fraction is 0.269, the molality is 4.71 mol/kg, and the molarity is 2.94 M.