What is the concentration of a solution made by dissolving 0.99mol of acetic acid in enough water to make 0.58L of solution?

[CH₃COOH] = 1.70 M

Explanation:

When we talk about concentration we can determine Molarity

Molarity determines the moles of solute that are contained in 1L of solution.

In this case our solute is the acetic acid.

M = mol/L

M = 0.99 mol /0.58L → 1.70 M

We can also make a rule of three

In 0.58 L of solution we have 0.99 moles of solute

In 1 L of solution we may have (1 . 0.99) / 0.58 = 1.70 moles

Acetic acid is a weak acid, partially dissociated in water.

CH₃COOH  +  H₂O  ⇄  CH₃COO⁻  +  H₃O⁺     Ka

did anyone actually do this? like if anyone has submitted this i rly need .