Molality of $$0.8$$ M $$H_2SO_4$$ solution (density $$1.06$$ g cm$$^{-3}$$) is ______ $$\times 10^{-3}$$ m.

The problem is to find the molality of a $$0.8$$ M $$H_2SO_4$$ solution whose density is $$1.06$$ g/cm$$^3$$; for $$H_2SO_4$$, the molar mass is $$2(1) + 32 + 4(16) = 98$$ g/mol.

Considering 1 litre (1000 mL) of this solution, the total mass is density × volume = $$1.06 \times 1000 = 1060$$ g.

With 0.8 mol of $$H_2SO_4$$ per litre, the mass of the solute is $$0.8 \times 98 = 78.4$$ g.

The mass of the solvent is then $$1060 - 78.4 = 981.6$$ g, or $$0.9816$$ kg.

The molality is given by $$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$$, so $$m = \frac{0.8}{0.9816} = 0.8150 \text{ mol/kg}$$.

Expressing in the required form: $$m = 815 \times 10^{-3}$$ m, which gives a numerical value of 815.