The density of toluene \(\left(\mathrm{C}_{7} \mathrm{H}_{8}\right)\) is \(0.867 \mathrm{~g} / \mathrm{mL},\) and the density of thiophene \(\left(\mathrm{C}_{4} \mathrm{H}_{4} \mathrm{~S}\right)\) is \(1.065 \mathrm{~g} / \mathrm{mL}\). A solution is made by dissolving \(8.10 \mathrm{~g}\) of thiophene in \(250.0 \mathrm{~mL}\) of toluene. (a) Calculate the mole fraction of thiophene in the solution. (b) Calculate the molality of thiophene in the solution. (c) Assuming that the volumes of the solute and solvent are additive, what is the molarity of thiophene in the solution?

First, we need to find the number of moles of thiophene and toluene in the solution. To do this, we'll use the mass of thiophene and the volume of toluene given, along with their respective densities and molar masses. Molar mass of thiophene (C4H4S): \(4(12.01 \mathrm{~g/mol}) + 4(1.008 \mathrm{~g/mol}) + 32.07 \mathrm{~g/mol} = 84.14 \mathrm{~g/mol}\) Molar mass of toluene (C7H8): \(7(12.01 \mathrm{~g/mol}) + 8(1.008 \mathrm{~g/mol}) = 92.14 \mathrm{~g/mol}\) Now, we can find the number of moles of each. Moles of thiophene: \(\frac{8.10 \mathrm{~g}}{84.14 \mathrm{~g/mol}} = 0.09629 \mathrm{~mol}\) Mass of toluene: \(250.0 \mathrm{~mL} \times 0.867 \mathrm{~g/mL} = 216.75 \mathrm{~g}\) Moles of toluene: \(\frac{216.75 \mathrm{~g}}{92.14 \mathrm{~g/mol}} = 2.352 \mathrm{~mol}\)

Now that we have the number of moles of both thiophene and toluene, we can calculate the mole fraction of thiophene in the solution using the formula: Mole fraction of thiophene = \(\frac{\text{moles of thiophene}}{\text{moles of thiophene} + \text{moles of toluene}}\) Mole fraction of thiophene = \(\frac{0.09629}{0.09629 + 2.352} = 0.03931\)

To calculate the molality of thiophene in the solution, we need to divide the number of moles of thiophene by the mass of toluene in kilograms: Molality of thiophene = \(\frac{\text{moles of thiophene}}{\text{mass of toluene in kg}}\) Molality of thiophene = \(\frac{0.09629}{0.21675} = 0.444 \mathrm{~mol/kg}\)

Assuming the volumes of solute (thiophene) and solvent (toluene) are additive, we can calculate the total volume of the solution: Volume of thiophene: \(\frac{8.10 \mathrm{~g}}{1.065 \mathrm{~g/mL}} = 7.606 \mathrm{~mL}\) Total solution volume: \(250.0 \mathrm{~mL} + 7.606 \mathrm{~mL} = 257.6 \mathrm{~mL}\) Now, we can calculate the molarity of thiophene using the formula: Molarity of thiophene = \(\frac{\text{moles of thiophene}}{\text{volume of solution in L}}\) Molarity of thiophene = \(\frac{0.09629}{0.2576} = 0.374 \mathrm{~M}\) The answers for each part of the exercise are as follows: a) Mole fraction of thiophene = 0.03931 b) Molality of thiophene = 0.444 mol/kg c) Molarity of thiophene = 0.374 M