An aqueous antifreeze solution is \(40.0 \%\) ethylene glycol \(\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\right)\) by mass. The density of the solution is \(1.05 \mathrm{~g} / \mathrm{cm}^{3}\). Calculate the molality, molarity, and mole fraction of the ethylene glycol.

Let's consider a 100 g of the solution. Given that the solution is 40% ethylene glycol by mass, this means we have 40 g of ethylene glycol and 60 g of water in the 100 g solution.

To calculate moles, we can use the formula: moles = mass / molar_mass The molar mass of ethylene glycol (C2H6O2) is 62.07 g/mol and the molar mass of water (H2O) is 18.015 g/mol. Moles of ethylene glycol = \( \frac{40 g}{62.07 g/mol} = 0.644 mol \) Moles of water = \( \frac{60 g}{18.015 g/mol} = 3.331 mol \)

Molality formula: Molality = \( \frac{moles\_of\_solute}{mass\_of\_solvent(kg)} \) Molality of ethylene glycol = \( \frac{0.644 mol}{0.060 kg} = 10.73 \, mol/kg \) Molarity formula: Molarity = \( \frac{moles\_of\_solute}{volume\_of\_solution(L)} \) To find the volume, we can use the formula: volume = \( \frac{mass}{density} \) The mass of the solution is 100 g, and the density is 1.05 g/cm³ or 1.05 g/mL. Volume of the solution = \( \frac{100 g}{1.05 g/mL} = 95.24 mL = 0.09524 L \) Molarity of ethylene glycol = \( \frac{0.644 mol}{0.09524 L} = 6.764 M \) Mole fraction formula: Mole fraction = \( \frac{moles\_of\_component}{total\_moles} \) Total moles = moles of ethylene glycol + moles of water = 0.644 mol + 3.331 mol = 3.975 mol Mole fraction of ethylene glycol = \( \frac{0.644 mol}{3.975 mol} = 0.162 \) Now we have the calculated values for molality, molarity, and mole fraction of ethylene glycol in the solution: Molality: 10.73 mol/kg Molarity: 6.764 M Mole Fraction: 0.162