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

Let’s assume we have 100 mL of the solution. Since it’s a 10% \(v/v\) solution, the volume of \(\text{Br}_2\) is 10 mL and the volume of \(\text{CCl}_4\) is 90 mL.
Step 1: Calculate the mass of \(\text{Br}_2\)
\[\text{Mass of \(\text{Br}_2\)} = \text{Volume} \times \text{Density} = 10 \, \text{mL} \times 3.2 \, \text{g/mL} = 32 \, \text{g}.\]
\[\text{Moles of \(\text{Br}_2\)} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{32 \, \text{g}}{160 \, \text{g/mol}} = 0.2 \, \text{mol}.\]
Step 2: Calculate the mass of \(\text{CCl}_4\)
\[\text{Mass of \(\text{CCl}_4\)} = \text{Volume} \times \text{Density} = 90 \, \text{mL} \times 1.6 \, \text{g/mL} = 144 \, \text{g}.\]
\[\text{Molar mass of \(\text{CCl}_4\)} = 12 + (4 \times 35.5) = 12 + 142 = 154 \, \text{g/mol}.\]
Step 3: Calculate the molality (\(m\))
\[\text{Molality} (m) = \frac{\text{Moles of solute}}{\text{Mass of solvent (in kg)}}.\]
\[m = \frac{0.2 \, \text{mol}}{144 \, \text{g} / 1000 \, \text{g/kg}} = \frac{0.2 \, \text{mol}}{0.144 \, \text{kg}} = 1.3888 \, \text{mol/kg}.\]
\[m \approx 1.39 \, \text{mol/kg}.\]
Final Answer:
\[x = 139.\]