The molality of a $$10\%(v/V)$$ solution of di-bromine solution in CCl$$_4$$ (carbon tetrachloride) is x. $$x = $$ ______ $$\times 10^{-2}$$ M. (Nearest integer)
[Given: molar mass of Br$$_2 = 160$$ g mol$$^{-1}$$, atomic mass of C $$= 12$$ g mol$$^{-1}$$, atomic mass of Cl $$= 35.5$$ g mol$$^{-1}$$, density of dibromine $$= 3.2$$ g cm$$^{-3}$$, density of CCl$$_4 = 1.6$$ g cm$$^{-3}$$]

We need to find the molality of a $$10\%$$ (v/V) solution of dibromine (Br$$_2$$) in CCl$$_4$$.

Interpret 10% (v/V).

$$10\%$$ (v/V) means 10 mL of Br$$_2$$ is present in 100 mL of solution.

Find the mass of Br$$_2$$ (solute).

Mass of Br$$_2$$ = Volume $$\times$$ Density = $$10 \times 3.2 = 32$$ g

Moles of Br$$_2$$ = $$\frac{32}{160} = 0.2$$ mol

Find the mass of CCl$$_4$$ (solvent).

Volume of CCl$$_4$$ = Total volume $$-$$ Volume of Br$$_2$$ = $$100 - 10 = 90$$ mL

Mass of CCl$$_4$$ = $$90 \times 1.6 = 144$$ g = $$0.144$$ kg

Calculate molality.

$$ \text{Molality} = \frac{\text{Moles of solute}}{\text{Mass of solvent (kg)}} = \frac{0.2}{0.144} = 1.3889 \text{ mol/kg} $$

Express in the required form.

$$x \times 10^{-2} = 1.3889$$

$$x = 138.89 \approx 139$$ (nearest integer)

The answer is $$\boxed{139}$$.