A 6.50 molal solution of KOH (aq.) has a density of 1.89 g cm$$^{-3}$$. The molarity of the solution is ________ mol dm$$^{-3}$$. (Round off to the Nearest Integer).
[Atomic masses: K: 39.0 u; O: 16.0 u; H: 1.0 u]

We are given a 6.50 molal solution of KOH with a density of 1.89 g/cm$$^3$$. We need to find the molarity of the solution.

A 6.50 molal solution means 6.50 moles of KOH are dissolved in 1000 g (1 kg) of water (solvent). The molar mass of KOH is $$39 + 16 + 1 = 56$$ g/mol.

The mass of KOH in this solution is $$6.50 \times 56 = 364$$ g.

The total mass of the solution is $$1000 + 364 = 1364$$ g.

Using the density, the volume of the solution is $$V = \frac{1364}{1.89} = 721.69 \text{ cm}^3 = 0.72169 \text{ dm}^3$$.

The molarity is the number of moles of solute per dm$$^3$$ of solution: $$M = \frac{6.50}{0.72169} = 9.007$$ mol dm$$^{-3}$$.

Rounding off to the nearest integer, the molarity is $$\mathbf{9}$$ mol dm$$^{-3}$$.