Given information:
- Density of 2M aqueous solution of NaOH = 1.28g/cm3

To find:
- Molality of the solution

Formula:
- Molality (m) = moles of solute / mass of solvent in kg

Step-by-step solution:
1. Calculate the molar mass of NaOH:
- Na = 23 g/mol
- O = 16 g/mol
- H = 1 g/mol
- Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol

2. Calculate the mass of NaOH present in 1 liter (1000 ml) of 2M solution:
- 2M solution means there are 2 moles of NaOH in 1 liter of solution
- Mass of 2 moles of NaOH = 2 x 40 = 80 g
- Density of the solution = 1.28 g/cm3
- Volume of 80 g of solution = 80 / 1.28 = 62.5 cm3 = 0.0625 L
- So, 1 liter of 2M solution contains 0.0625 x 16 = 1 g/cm3 of NaOH

3. Calculate the mass of water present in 1 liter of solution:
- Mass of 1 liter of solution = 1000 g
- Mass of NaOH in 1 liter of solution = 1 g
- Mass of water in 1 liter of solution = 1000 - 1 = 999 g

4. Calculate the molality of the solution:
- Moles of NaOH in 1 kg of water = 1 / 40 = 0.025 mol
- Mass of water in 1 kg of solution = 1000 - 1280 = 720 g = 0.72 kg
- Molality of the solution = 0.025 / 0.72 = 0.0347 m = 1.34 m (approx.)

Answer:
The molality of the solution is 1.34m (approx.).