The fraction of acetic acid that is dissociated is 0.18
Why?
The chemical equation for the dissociation of acetic acid (HAc) is the following:
HAc(aq) + H₂O(l) ⇄ H₃O⁺(aq) + Ac⁻(aq)
To find the fraction of acetic acid that is in the dissociated form (f), we apply the following equation (Ka for acetic acid is 1.76*10⁻⁵). This equation comes from solving the equation of the equilibrium constant for the dissociated fraction of HAc:

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<span>the highest ionization energy is Be, it is higher and righter than other elements in periodic table, so it harder to remove electron from its atom</span>
Assume it is 1 litre and weighs 1kg.
2 percent of 1 kg is 20g.
20g divided by molar mass of NaOH.
20g divide by 40 = 0.5 mole
0.5 mole in a litre would be 0.5M
That is the answer: 0.5M