Given data

• Molarity of acetic acid solution: $M = 2.05\,\text{mol L}^{-1}$
• Density of solution: $\rho = 1.02\,\text{g mL}^{-1}$
• We need to find molality.

Take 1 L of solution Since molarity is given, let us consider $1\,\text{L}$ of solution.

Then, moles of acetic acid in $1\,\text{L}$ solution: $n_{\text{solute}} = 2.05\,\text{mol}$

Find mass of 1 L solution Density $= 1.02\,\text{g mL}^{-1}$

So mass of $1000\,\text{mL}$ solution is: $m_{\text{solution}} = 1.02 \times 1000 = 1020\,\text{g}$

Find mass of acetic acid (solute) Molar mass of acetic acid, $\text{CH}_3\text{COOH}$: $M_r = 2(12) + 4(1) + 2(16) = 60\,\text{g mol}^{-1}$

Therefore mass of solute: $m_{\text{solute}} = 2.05 \times 60 = 123\,\text{g}$

Find mass of solvent (water) $m_{\text{solvent}} = m_{\text{solution}} - m_{\text{solute}}$ $m_{\text{solvent}} = 1020 - 123 = 897\,\text{g} = 0.897\,\text{kg}$

Calculate molality Molality is moles of solute per kg of solvent: $m = \frac{n_{\text{solute}}}{\text{mass of solvent in kg}}$ $m = \frac{2.05}{0.897}$ $m \approx 2.28\,\text{mol kg}^{-1}$

Match with options $\boxed{2.28\,\text{mol kg}^{-1}}$ So the correct option is A.

Comparison with stored answer Stored correct answer: A

My derived answer also gives A, so they agree.