Molarity (M) of an aqueous solution containing $$x$$ g of anhyd. $$CuSO_4$$ in $$500$$ mL solution at $$32°C$$ is $$2 \times 10^{-1}$$ M. Its molality will be ______ $$\times 10^{-3}$$ m. (nearest integer). [Given density of the solution $$= 1.25$$ g/mL]

Given: Molarity = 0.2 M, volume = 500 mL, density = 1.25 g/mL.

Moles of $$CuSO_4$$ = $$0.2 \times 0.5 = 0.1$$ mol.

Molar mass of anhydrous $$CuSO_4$$ = 64 + 32 + 64 = 160 g/mol.

Mass of $$CuSO_4$$ = $$0.1 \times 160 = 16$$ g.

Mass of solution = $$500 \times 1.25 = 625$$ g.

Mass of solvent (water) = $$625 - 16 = 609$$ g = 0.609 kg.

Molality = $$\frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.1}{0.609} = 0.16420$$ m.

Converting: $$0.16420$$ m = $$164.2 \times 10^{-3}$$ m.

Nearest integer: $$164 \times 10^{-3}$$ m.

The answer is $$\boxed{164}$$.