Density of 2.03 M aqueous solution of acetic acid is 1.017 g mL-1 molecular mass of acetic acid is 60. Calculatethe molality of solution?(1) 2.27(2) 1.27(3) 3.27(4) 4.27

Density of 2.03 M aqueous solution of acetic acid is 1.017 g mL-1 mole

1.	Density of 2.03 M aqueous solution of acetic acid is 1.017 g mL-1 molecular mass of acetic acid is 60. Calculatethe molality of solution?(1) 2.27(2) 1.27(3) 3.27(4) 4.27
Answer» Given: Molarity of solution = 2.03 MDensity of solution = 1.017 g mL-1Molar mass of acetic acid = 60 g mol-1To find : Molality of solutionSolve:Molality = Number of moles / Mass of solvent in kgWe have the number of moles but we need to find the mass of solvent.As we know molarity is number of moles / Volume of solution (in liters)Molarity = Number of moles if we have 1 L solution.Number of moles = 2.03 in 1 L or 1000 mL solutionDensity = Mass / VolumeMass = DensityVolumeMass = 1.0171000= 1017 gMass of solution = 1017 gMass of acetic acid in solution = Number of molesMolar mass= 2.0360= 121.8 gMass of solvent = mass of solution - mass of acetic acid= 1017-121.8= 895.2 g or 895.210^-3kgMolality = Number of moles / Mass of solvent (in kg)2.03 / 895.210^-3)= 2.27 m