JEE Chemistry 2023 Question with Solution

Given: A $10\%$ $\text{(v/v)}$ solution of Br_2 in CCl_4.

Find: The value of $x$ in $x \times 10^{-2}$.

Assume $100 \, \text{mL}$ of solution. Then volume of $Br_2$ is $10 \, \text{mL}$ and volume of $CCl_4$ is $90 \, \text{mL}$.

Step 1: Calculate the mass of $Br_2$

$$\text{Mass of } Br_2 = \text{Volume} \times \text{Density} = 10 \, \text{mL} \times 3.2 \, \text{g/mL} = 32 \, \text{g}$$ $$\text{Moles of } Br_2 = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{32 \, \text{g}}{160 \, \text{g/mol}} = 0.2 \, \text{mol}$$

Step 2: Calculate the mass of $CCl_4$

$$\text{Mass of } CCl_4 = \text{Volume} \times \text{Density} = 90 \, \text{mL} \times 1.6 \, \text{g/mL} = 144 \, \text{g}$$ $$\text{Molar mass of } CCl_4 = 12 + (4 \times 35.5) = 12 + 142 = 154 \, \text{g/mol}$$

Step 3: Calculate the molality

$$\text{Molality } (m) = \frac{\text{Moles of solute}}{\text{Mass of solvent (in kg)}}$$ $$m = \frac{0.2 \, \text{mol}}{144 \, \text{g} / 1000 \, \text{g/kg}} = \frac{0.2 \, \text{mol}}{0.144 \, \text{kg}} = 1.3888 \, \text{mol/kg}$$ $$m \approx 1.39 \, \text{mol/kg}$$

So,

$$x = 139$$

Therefore, the final answer is $139$.

Given: $10\%$ $\text{(v/v)}$ means $10 \, \text{mL}$ solute per $100 \, \text{mL}$ solution.

Find: Convert the computed molality into the form $x \times 10^{-2}$.

Take $100 \, \text{mL}$ solution.

• $Br_2$ volume = $10 \, \text{mL}$
• $CCl_4$ volume = $90 \, \text{mL}$

Using the given densities:

$$\text{Mass of } Br_2 = 10 \times 3.2 = 32 \, \text{g}$$ $$\text{Mass of } CCl_4 = 90 \times 1.6 = 144 \, \text{g}$$

Now calculate moles of $Br_2$:

$$\text{Moles of } Br_2 = \frac{32}{160} = 0.2$$

Convert solvent mass into kilograms:

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

Then molality is

$$m = \frac{0.2}{0.144} = 1.3888 \approx 1.39$$

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

Therefore, $x = 139$.