An aqueous KCl solution of density 1.20 g mL$$^{-1}$$ has a molality of 3.30 mol kg$$^{-1}$$. The molarity of the solution in mol L$$^{-1}$$ is _________ (The Nearest integer)
[Molar mass of KCl = 74.5 g]

We have an aqueous solution of potassium chloride with density $$d = 1.20 \text{ g mL}^{-1}$$ and molality $$m = 3.30 \text{ mol kg}^{-1}$$. Our task is to convert this molality into molarity $$M$$, remembering that

$$\boxed{M = \dfrac{\text{moles of solute}}{\text{volume of solution in litres}}}$$

To connect molality with molarity, it is convenient to begin with a hypothetical sample containing exactly $$1.00\ \text{kg}$$ of the solvent (water). This choice is made because the definition of molality is

$$m = \dfrac{n_{\text{solute}}}{\text{mass of solvent in kg}}.$$

Substituting the given value $$m = 3.30$$ and the chosen solvent mass $$1.00$$ kg, we obtain the moles of KCl present:

$$n_{\text{KCl}} = m \times (\text{mass of solvent in kg}) = 3.30 \times 1.00 = 3.30\ \text{mol}.$$

Now we convert these moles into grams using the molar mass of KCl, $$M_{r}(\text{KCl}) = 74.5\ \text{g mol}^{-1}$$:

$$\text{mass of KCl} = n_{\text{KCl}} \times M_{r} = 3.30 \times 74.5 = 245.85\ \text{g}.$$

The total mass of the resulting solution is therefore

$$\text{mass of solution} = \text{mass of solvent} + \text{mass of solute} = 1000\ \text{g} + 245.85\ \text{g} = 1245.85\ \text{g}.$$

Using the density formula $$\rho = \dfrac{\text{mass}}{\text{volume}}$$, the volume of this solution is

$$V = \dfrac{\text{mass of solution}}{\rho} = \dfrac{1245.85\ \text{g}}{1.20\ \text{g mL}^{-1}} = 1038.208\ \text{mL}.$$

Converting millilitres to litres, we get

$$V = 1038.208\ \text{mL} = 1.038208\ \text{L}.$$

Finally, we calculate the molarity:

$$M = \dfrac{n_{\text{KCl}}}{V} = \dfrac{3.30\ \text{mol}}{1.038208\ \text{L}} = 3.18\ \text{mol L}^{-1}.$$

Rounding this value to the nearest integer gives

$$M \approx 3.$$

So, the answer is $$3$$.