Question

in this question we have to do solve the following equations I know answer but I want step by step solutions the answers are given last of the book answer of (g) 4 , (h) 0.476 , (I) 6/7 so all answer are here please tell me step by step solutions like I take picture of the questions like this way​

Answer 1

Answer:

A solution of acetic acid (CH3COOH), methanol (CH3OH), and water is prepared such that the mole fraction of acetic acid is 0.01335, and the mole fraction of methanol is 0.03288. The density of the solution is measured to be 0.9889 g/mL. What is the molarity (in M) of acetic acid in this solutionA solution of acetic acid (CH3COOH), methanol (CH3OH), and water is prepared such that the mole fraction of acetic acid is 0.01335, and the mole fraction of methanol is 0.03288. The density of the solution is measured to be 0.9889 g/mL. What is the molarity (in M) of acetic acid in this solutionA solution of acetic acid (CH3COOH), methanol (CH3OH), and water is prepared such that the mole fraction of acetic acid is 0.01335, and the mole fraction of methanol is 0.03288. The density of the solution is measured to be 0.9889 g/mL. What is the molarity (in M) of acetic acid in this solutionA solution of acetic acid (CH3COOH), methanol (CH3OH), and water is prepared such that the mole fraction of acetic acid is 0.01335, and the mole fraction of methanol is 0.03288. The density of the solution is measured to be 0.9889 g/mL. What is the molarity (in M) of acetic acid in this solutionA solution of acetic acid (CH3COOH), methanol (CH3OH), and water is prepared such that the mole fraction of acetic acid is 0.01335, and the mole fraction of methanol is 0.03288. The density of the solution is measured to be 0.9889 g/mL. What is the molarity (in M) of acetic acid in this solution