Calculate the molarity and mole fraction of acetone in a 1.00 -m solution of acetone \(\left(\mathrm{CH}_{3} \mathrm{COCH}_{3}\right)\) in ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) . (Density of acetone \(=0.788 \mathrm{g} / \mathrm{cm}^{3} ;\) density of ethanol \(=\) 0.789 \(\mathrm{g} / \mathrm{cm}^{3} .\) ) Assume that the volumes of acetone and ethanol add.

: We know the molality (m) of the solution is 1.00 mol/kg. The molality definition is: \(molality = \frac{moles_{solute}}{mass_{solvent}(kg)}\) We are given a molality of 1.00 mol/kg, so: \(1.00 mol/kg = \frac{moles_{acetone}}{mass_{ethanol}(kg)}\) Now, we need to calculate the mass of the solute (acetone) and the solvent (ethanol).

: Since the molality is 1.00 mol/kg, it means there is 1 mole of acetone for 1 kg of ethanol: \(moles_{acetone} = 1.00 mol\)

: To calculate the mass of the acetone and ethanol, we use the density and volume relationship: \(Density = \frac{mass}{volume}\) To find the mass, we first need the volume of the solute and solvent. Since their volumes are additive, the total volume is given as: \(V_{total} = V_{acetone} + V_{ethanol}\) But we don't have the total volume given. So we first calculate the mass of ethanol and acetone with the data we do have. Mass of acetone can be calculated from moles and molar mass: \(Molar\, Mass\, of\,acetone = (12.01 * 3) + (1.01 * 6) + (16.00) = 58.08 g/mol\) \(mass_{acetone}= moles_{acetone} * Molar\,mass_{acetone}\) \(mass_{acetone} = 1.00 * 58.08\,g\) Since there is 1 kg of ethanol, we can convert it to grams: \(mass_{ethanol} = (1.00\, kg) * (1000\, g/kg) = 1000\,g\)

: Using densities, we calculate the volume of acetone and ethanol: \(volume_{acetone} = \frac{mass_{acetone}}{density_{acetone}}\) \(volume_{acetone} = \frac{58.08 g}{0.788 g/cm^3} = 73.68 cm^3 = 0.07368 dm^3\) \(volume_{ethanol} = \frac{mass_{ethanol}}{density_{ethanol}}\) \(volume_{ethanol} = \frac{1000 g}{0.789 g/cm^3} = 1267.31 cm^3 = 1.26731 dm^3\)

: Now that we have the volumes of acetone and ethanol, we can determine the total volume: \(V_{total} = V_{acetone} + V_{ethanol}\) \(V_{total}= 0.07368\, dm^3 + 1.26731\, dm^3\) \(V_{total} = 1.34099\, dm^3\)

: Finally, we can calculate the molarity of acetone using its moles and the total volume of the solution: \(Molarity = \frac{moles_{solute}}{volume_{solution}(dm^3)}\) \(Molarity = \frac{1.00\, mol}{1.34099\, dm^3}\) \(Molarity = 0.745\, M\)

: The mole fraction of acetone in the solution can be calculated by: \(Mole Fraction = \frac{moles_{acetone}}{moles_{acetone}+ moles_{ethanol}}\) We already have the moles of acetone, and we can calculate the moles of ethanol using the mass of ethanol and its molar mass: \(Molar\, mass_{ethanol}=(12.01*2)+ (1.01*6) +(16.00)=46.07\,g/mol\) \(moles_{ethanol} = \frac{mass_{ethanol}}{molar\, mass_{ethanol}}\) \(moles_{ethanol} = \frac{1000\, g}{46.07\, g/mol} = 21.707\, mol\) Now, we can find the mole fraction of acetone: \(Mole Fraction = \frac{1.00\, mol}{1.00\, mol + 21.707\, mol}\) \(Mole Fraction = 0.044\) Summary: The molarity of acetone in the solution is 0.745 M and the mole fraction of acetone in the solution is 0.044.