The density of 3M solution of NaCl is 1.0 g mL$$^{-1}$$. Molality of the solution is _____ $$\times 10^{-2}$$ m (Nearest integer).
Given: Molar mass of Na is 23 and Cl is 35.5 g mol$$^{-1}$$ respectively.

We have a 3M solution of NaCl with density = 1.0 g/mL. The molar mass of NaCl = 23 + 35.5 = 58.5 g/mol.

In 1 litre of solution, the mass of solution = 1000 mL $$\times$$ 1.0 g/mL = 1000 g, and the moles of NaCl = 3 mol, giving mass of NaCl = 3 $$\times$$ 58.5 = 175.5 g.

Now, the mass of solvent (water) is:

$$m_{water} = 1000 - 175.5 = 824.5 \text{ g} = 0.8245 \text{ kg}$$

Molality is defined as moles of solute per kg of solvent:

$$\text{Molality} = \frac{3}{0.8245} = 3.6386 \text{ m}$$

In terms of $$\times 10^{-2}$$ m:

$$\text{Molality} = 363.86 \times 10^{-2} \text{ m} \approx 364 \times 10^{-2} \text{ m}$$

So, the answer is $$364$$.