Question

The molality of a 10% (v/v) solution of di-bromine solution in \(\text{CCl}_4\) (carbon tetrachloride) is \(x\). \(x = \, \_\_\_\_\ \times 10^{-2} \, \text{M}\). (Nearest integer)
Given:
Molar mass of \(\text{Br}_2 = 160 \, \text{g mol}^{-1}\)
Atomic mass of \(\text{C} = 12 \, \text{g mol}^{-1}\)
Atomic mass of \(\text{Cl} = 35.5 \, \text{g mol}^{-1}\)
Density of dibromine = \(3.2 \, \text{g cm}^{-3}\)
Density of \(\text{CCl}_4 = 1.6 \, \text{g cm}^{-3}\)

Hint:

Remember the formula for molality: Molality (m) = Moles of solute / Mass of solvent (in kg). Pay close attention to the units provided and required for the final answer. A v/v percentage means volume of solute per volume of solution.

Answer 1

Correct Answer: 139

To find the molality of the 10% (v/v) solution of bromine in \(\text{CCl}_4\), follow these steps:

Step 1: Calculate volume of \(\text{Br}_2\)
10% (v/v) implies 10 mL of \(\text{Br}_2\) is present in 100 mL of solution.

Step 2: Calculate mass of \(\text{Br}_2\)
Given density, \(\rho_{\text{Br}_2} = 3.2 \, \text{g cm}^{-3}\). Mass = Volume × Density.
Mass of \(\text{Br}_2 = 10 \, \text{cm}^{3} \times 3.2 \, \text{g cm}^{-3} = 32 \, \text{g}\).

Step 3: Calculate mass of \(\text{CCl}_4\)
Density of \(\text{CCl}_4 = 1.6 \, \text{g cm}^{-3}\).
Volume of \(\text{CCl}_4 = 100 - 10 = 90 \, \text{cm}^{3}\).
Mass of \(\text{CCl}_4 = 90 \, \text{cm}^{3} \times 1.6 \, \text{g cm}^{-3} = 144 \, \text{g}\).

Step 4: Calculate molality
Molality \((m) = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}\).
Moles of \(\text{Br}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{32 \,\text{g}}{160 \, \text{g mol}^{-1}} = 0.2 \, \text{mol}\).
Mass of solvent (\(\text{CCl}_4\)) in kg = \(0.144 \, \text{kg}\).
Molality \(m = \frac{0.2 \, \text{mol}}{0.144 \, \text{kg}} \approx 1.389 \, \text{mol kg}^{-1}\).

Step 5: Express in given format
As, \(x = 1.389 \times 10^{-2} \, \text{M}\), nearest integer is \(139\).

The computed value, \(139\), falls within the expected range (139,139).