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 take 100 grams of the solution, as it will make calculations easier (since it's 40% ethylene glycol). In 100 g of solution, we have: - 40 g of ethylene glycol (C2H6O2) - 60 g of water (as the solvent) To calculate molality, we need moles of solute (ethylene glycol) and mass of solvent (water) in kilograms: 1. Calculate the molar mass of ethylene glycol: C2H6O2: \(2\times12.01 + 6\times1.01 + 2\times16.00 = 62.07 \, g/mol\) 2. Calculate the moles of ethylene glycol in 40 g: \(moles = \frac{mass}{molar\:mass} = \frac{40\:g}{62.07\:\frac{g}{mol}} = 0.6445 \, mol\) 3. Convert the mass of water to kg: \(60\,g = 0.060\, kg\) 4. Calculate the molality: \(molality = \frac{moles\:of\: solute}{kg \:of\: solvent} = \frac{0.6445\:mol}{0.060\:kg} = 10.74 \, mol/kg\) So, the molality of ethylene glycol in the solution is 10.74 mol/kg.

Now, we need to find the molarity of ethylene glycol, which is the concentration in moles of solute per liter of solution. We know that 100 grams of the solution has the density of 1.05 g/cm³. 1. Convert the density to g/mL: \(1.05 \frac{g}{cm^3} = 1.05 \frac{g}{mL}\) 2. Calculate the volume of the 100 g solution: \(volume = \frac{mass}{density} = \frac{100\:g}{1.05\:\frac{g}{mL}} = 95.24\, mL\) 3. Convert the volume to liters: \(95.24\:mL = 0.09524\, L\) 4. Calculate the molarity: \(molarity = \frac{moles\:of\:solute}{L\:of\:solution} = \frac{0.6445\:mol}{0.09524\:L} = 6.77\, M\) So, the molarity of ethylene glycol in the solution is 6.77 M.

Finally, we need to find the mole fraction of ethylene glycol. The mole fraction is the ratio of the number of moles of ethylene glycol to the total number of moles in the solution. 1. Calculate the moles of water in the solution: Molar mass of water (H2O): \(18.015 \, g/mol\) \(moles = \frac{mass}{molar\:mass} = \frac{60\:g}{18.015\:\frac{g}{mol}} = 3.3314 \, mol\) 2. Calculate the mole fraction of ethylene glycol: \(mole\:fraction = \frac{moles\: of\: ethylene\: glycol}{total\: moles} = \frac{0.6445\:mol}{0.6445\:mol + 3.3314\:mol} = 0.1621\) So, the mole fraction of ethylene glycol in the solution is 0.1621. In summary, the molality of the ethylene glycol in the antifreeze solution is 10.74 mol/kg, the molarity is 6.77 M, and the mole fraction is 0.1621.