Question

(65-(684÷(37+65) find the value of ​

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 solution