To determine the Henry's law constant for bromobenzene, we use the relationship provided by Henry's Law, which states that the vapor pressure of a solute above a dilute solution is directly proportional to its mole fraction in the solution. Mathematically, Henry's Law is given by:

\(P = k_H \cdot x\)

where:

• \(P\) is the vapor pressure of the solute (bromobenzene in this case) over the solution.
• \(k_H\) is the Henry's law constant.
• \(x\) is the mole fraction of the solute in the solution.

Given:

• Vapor pressure of bromobenzene, \(P = 24.0 \, \text{kPa}\)
• Molality of the solution is \(0.1 \, \text{mol} \, \text{kg}^{-1}\)

For a dilute solution, the mole fraction \(x\) can be approximated as the molality when the solvent is water. Since water has a molar mass of approximately 18 g/mol, the mole fraction \(x\) is:

\(x \approx \frac{0.1}{55.5} \, \text{since} \, 1 \, \text{kg of water} \approx 55.5 \, \text{mol of water}\)

Thus, the mole fraction \(x = \frac{0.1}{55.5} = 0.0018\).

Substituting \(P\) and \(x\) into the equation for Henry's Law:

\(24.0 = k_H \cdot 0.0018\)

Solving for \(k_H\):

\(k_H = \frac{24.0}{0.0018} = 24000 \, \text{kPa} \cdot \text{mol}^{-1}\)

However, the unit needs to be expressed in terms of kPa kg mol^-1. Hence, dividing by 1000 to convert from mol^-1 to kg mol^-1, we get:

\(k_H = 240 \, \text{kPa} \cdot \text{kg} \cdot \text{mol}^{-1}\)

Therefore, the Henry's law constant of bromobenzene is 240 kPa kg mol^-1. This matches the first option, which is \(240 \, \text{kPa} \, \text{kg} \, \text{mol}^{-1}\).