when 6g acetic acid is dissolved in 1000cm3 of solution then how many molecules ionize out of 1000 acetic acid molecules

24.8 molecules are ionized from 1000 acetic acid molecules.

Explanation:

Acetic acid, CH₃COOH dissociates in water, thus:

CH₃COOH ⇄ CH₃COO⁻ + H⁺

Ka = 6.3x10⁻⁵ = [CH₃COO⁻] [H⁺] / [CH₃COOH]

That means amount of CH₃COO⁻ (the dissociated form) that are produced is followed by the equilibrium of the weak acid.

The initial molar concentration of acetic acid (Molar mass: 60g/mol) is:

6g ₓ (1mol / 60g) = 0.1 moles acetic acid, in 1000cm³ = 1L.

0.1 moles / L = 0.1M

The 0.1M of acetic acid will dissociate producing X of CH₃COO⁻ and H⁺, thus:

[CH₃COOH] = 0.1M - X

[CH₃COO⁻] = X

[H⁺] = X

Replacing in Ka formula:

6.3x10⁻⁵ = [CH₃COO⁻] [H⁺] / [CH₃COOH]

6.3x10⁻⁵ = [X] [X] / [0.1 - X]

6.3x10⁻⁶ - 6.3x10⁻⁵X = X²

6.3x10⁻⁶ - 6.3x10⁻⁵X - X² = 0

Solving for X

X = - 0.0025 → False solution, there is no negative concentrations.

X = 0.00248M

That means, a 0.1M of acetic acid produce:

[CH₃COO⁻] = X = 0.00248M solution of the ionized form.

In a basis of 1000 molecules:

1000 molecules × (0.00248M / 0.1M) = 24.8

the maximum mass of nh3 is nh3 in grams