A solution is made containing \(50.0 \mathrm{~g}\) of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) in \(1000 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O} .\) Calculate \((\mathbf{a})\) the mole fraction of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH},\) (b) the mass percent of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\), (c) the molality of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\).

(To find the mole fraction, mass percent, and molality of ethanol in the solution, we first need to find the molar mass of ethanol and water. For ethanol, the molar mass will be the sum of the molar masses of 2 carbon atoms, 6 hydrogen atoms, and 1 oxygen atom. For water, the molar mass will be the sum of the molar masses of 2 hydrogen atoms and 1 oxygen atom.)

(Next, we need to calculate the moles of ethanol and water in the solution. We do this by dividing the mass of each substance by its molar mass.)

(We can find the mole fraction of ethanol by dividing the moles of ethanol by the total moles of both ethanol and water.)

(The mass percent of ethanol can be calculated by dividing the mass of ethanol in the solution by the total mass of the solution and multiplying the result by 100.)

(The molality of ethanol can be found by dividing the moles of ethanol by the mass of the solvent (water) in kilograms.) Now let's perform the calculations. Step 1: Molar Mass of Ethanol and Water Ethanol (\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\)): Molar mass = (2 × 12.01) + (6 × 1.01) + (1 × 16.00) = 46.07 g/mol Water (\(\mathrm{H}_{2} \mathrm{O}\)): Molar mass = (2 × 1.01) + (1 × 16.00) = 18.02 g/mol Step 2: Moles of Ethanol and Water Moles of ethanol = Mass of ethanol / Molar mass of ethanol = \(50.0 \mathrm{~g} / 46.07 \mathrm{~g/mol} = 1.085 \mathrm{~mol}\) Moles of water = Mass of water / Molar mass of water = \(1000 \mathrm{~g} / 18.02 \mathrm{~g/mol} = 55.49 \mathrm{~mol}\) Step 3: Mole Fraction of Ethanol Mole fraction of ethanol = Moles of ethanol / Total moles = \(1.085 \mathrm{~mol} / (1.085\mathrm{~mol} + 55.49 \mathrm{~mol}) \approx 0.019\) Step 4: Mass Percent of Ethanol Mass percent of ethanol = (Mass of ethanol / Total mass) × 100 = \((50.0 \mathrm{~g} / (50.0 \mathrm{~g} + 1000 \mathrm{~g})) × 100 = 4.76\%\) Step 5: Molality of Ethanol Molality of ethanol = Moles of ethanol / Mass of water in kg = \(1.085 \mathrm{~mol} / 1.000 \mathrm{~kg} = 1.085 \mathrm{~mol/kg}\) The results are as follows: (a) Mole fraction of ethanol: 0.019 (b) Mass percent of ethanol: 4.76% (c) Molality of ethanol: 1.085 mol/kg