Question

The density of a 2.0M Solution of acetic acid in water is 1.02 g / mL . calculate the mole fraction of acetic acid​

Answer 1

Answer:

We have a 2.05M solution of acetic acid, density 1.02g/ml

We have a 2.05M solution of acetic acid, density 1.02g/mlSo the mass of the solution is 1000 x 1.02 = 1020g/L

We have a 2.05M solution of acetic acid, density 1.02g/mlSo the mass of the solution is 1000 x 1.02 = 1020g/LThe molar mass of acetic acid is 60.05g/mol

We have a 2.05M solution of acetic acid, density 1.02g/mlSo the mass of the solution is 1000 x 1.02 = 1020g/LThe molar mass of acetic acid is 60.05g/mol60.05g/mol x 2.05mol = 123.1025g in solution

We have a 2.05M solution of acetic acid, density 1.02g/mlSo the mass of the solution is 1000 x 1.02 = 1020g/LThe molar mass of acetic acid is 60.05g/mol60.05g/mol x 2.05mol = 123.1025g in solutionMass of water in solution = 1020g – 123.1025 = 896.8975g

So 1kg of water would contain 2.05mol x 1000g/896.8975 = 2.286g of acetic acid

Therefore, the molality of this solution is 2.286

Answer 2

Molality =6.01²³ ×8⁷⁸⁰⁰⁰ = 2.28 mol kg−1.