1. Interpret the given concentration

A $10\%\,(v/v)$ solution means:

$\frac{\text{volume of solute}}{\text{volume of solution}}\times 100 = 10$

So, in $100\,\text{mL}$ of solution:

• Volume of $\mathrm{Br_2} = 10\,\text{mL}$
• Volume of $\mathrm{CCl_4} = 90\,\text{mL}$

2. Calculate mass of solute $\mathrm{Br_2}$

Given density of dibromine = $3.2\,\text{g mL}^{-1}$

$\text{mass of } \mathrm{Br_2} = 10\times 3.2 = 32\,\text{g}$

Moles of $\mathrm{Br_2}$:

$n_{\mathrm{Br_2}} = \frac{32}{160} = 0.2\,\text{mol}$

3. Calculate mass of solvent $\mathrm{CCl_4}$

Given density of $\mathrm{CCl_4} = 1.6\,\text{g mL}^{-1}$

$\text{mass of } \mathrm{CCl_4} = 90\times 1.6 = 144\,\text{g}$

Convert into kg:

$144\,\text{g} = 0.144\,\text{kg}$

4. Calculate molality

Molality is defined as:

$m = \frac{\text{moles of solute}}{\text{kg of solvent}}$

So,

$m = \frac{0.2}{0.144} = 1.3889\,\text{mol kg}^{-1}$

Thus,

$m \approx 1.39\,\text{m}$

5. Match with the required form

The question writes:

$x = \underline{\hspace{2cm}}\times 10^{-2}\,\mathrm{M}$

Numerically,

$1.39 = 139\times 10^{-2}$

Hence,

$x = 139$

6. Comparison with stored answer

Derived answer = $139$

Stored correct answer = $139$

So they agree.