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 $$Br_2 = 160 \text{ g mol}^{-1}$$, atomic mass of $$C = 12 \text{ g mol}^{-1}$$, atomic mass of $$Cl = 35.5 \text{ g mol}^{-1}$$, density of dibromine $$= 3.2 \text{ g cm}^{-3}$$, density of $$CCl_4 = 1.6 \text{ g cm}^{-3}$$]

A 10% (v/V) solution of $$Br_2$$ in $$CCl_4$$ implies that in 100 mL of solution there are 10 mL of $$Br_2$$ and consequently 90 mL of $$CCl_4$$.

Since the density of $$Br_2$$ is 3.2 g/cm³, the mass of the solute is $$\text{Mass of } Br_2 = 10 \text{ mL} \times 3.2 \text{ g/mL} = 32 \text{ g}$$, and thus the moles of solute are $$\text{Moles of } Br_2 = \frac{32}{160} = 0.2 \text{ mol}$$.

Substituting the density of $$CCl_4$$ (1.6 g/cm³) for the 90 mL solvent gives its mass as $$\text{Mass of } CCl_4 = 90 \text{ mL} \times 1.6 \text{ g/mL} = 144 \text{ g} = 0.144 \text{ kg}$$.

From this, the molality follows: $$\text{Molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.2}{0.144} = 1.389 \text{ m}$$.

This gives $$x = 1.389 = 139 \times 10^{-2}$$.

The answer is 139.