he density of a 2.05 M2.05 M solution of acetic acid in water is 1.02 g mL−11.02 g mL-1. The molality of the solution is

To find the molality of the solution, we need to know the moles of solute (acetic acid) and the mass of the solvent (water) in kilograms.

Step 1: Find the moles of acetic acid The molarity (M) is given as 2.05 M, which means there are 2.05 moles of acetic acid per liter of solution.

Step 2: Find the mass of the solution The density of the solution is given as 1.02 g/mL. To find the mass of 1 liter (1000 mL) of solution, multiply the density by the volume: 1.02 g/mL * 1000 mL = 1020 g.

Step 3: Find the mass of the acetic acid The molar mass of acetic acid (CH3COOH) is approximately 60.052 g/mol. Multiply the moles of acetic acid by its molar mass to find the mass: 2.05 mol * 60.052 g/mol = 123.1066 g.

Step 4: Find the mass of the water Subtract the mass of the acetic acid from the total mass of the solution to find the mass of the water: 1020 g - 123.1066 g = 896.8934 g. Convert this to kilograms by dividing by 1000: 0.8968934 kg.

Step 5: Calculate the molality Molality (m) is defined as the moles of solute per kilogram of solvent. Divide the moles of acetic acid by the mass of the water in kilograms to find the molality: 2.05 mol / 0.8968934 kg = 2.285 m.

So, the molality of the solution is approximately 2.285 mol/kg.