Step-by-step Solution:

1. Identify Given Information:

   • Mass of solute ($w_2$) = 2.5 g
   • Mass of solvent (water, $w_1$) = 100 g = 0.1 kg
   • Elevation in boiling point ($\\Delta T_b$) = 2 °C = 2 K
   • Ebullioscopic constant for water ($K_b$) = 0.76 K kg mol⁻¹
   • The question concerns the solution at 1 atm pressure. This implies the normal boiling point of pure water is 100 °C, and at this temperature, its vapour pressure ($P^o$) is 1 atm = 760 mm Hg.
   • We need to find the vapour pressure of the solution ($P_s$) at 100 °C.

2. State Relevant Formulas:

   • Elevation in Boiling Point: The relationship between the elevation in boiling point and the molality ($m$) of the solution is given by: $\Delta T_b = K_b \times m$
   • Raoult's Law (Relative Lowering of Vapour Pressure): The relative lowering of vapour pressure is equal to the mole fraction of the solute ($\\chi_2$): $\frac{P^o - P_s}{P^o} = \chi_2$

3. Use the Dilute Solution Approximation: The problem states to assume the concentration of the solute is much lower than the concentration of the solvent. For such dilute solutions, the mole fraction of the solute can be approximated as: $\chi_2 = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1}$ where $n_1$ and $n_2$ are the moles of the solvent and solute, respectively.

4. Relate the Two Colligative Properties: We can establish a direct relationship between the elevation in boiling point and the relative lowering of vapour pressure.

   • Molality ($m$) is defined as: $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{n_2}{w_1 (\text{kg})}$
   • The mass of the solvent can be expressed as $w_1(\text{kg}) = n_1 \times M_1(\text{kg})$, where $M_1$ is the molar mass of the solvent in kg/mol. For water, $M_1 = 18 \text{ g/mol} = 0.018 \text{ kg/mol}$.
   • Substituting this into the molality formula: $m = \frac{n_2}{n_1 \times M_1(\text{kg})} \implies \frac{n_2}{n_1} = m \times M_1(\text{kg})$
   • From the boiling point elevation formula, we have $m = \frac{\Delta T_b}{K_b}$.
   • Substituting this expression for $m$: $\frac{n_2}{n_1} = \frac{\Delta T_b}{K_b} \times M_1(\text{kg})$
   • Now, substitute this into the approximated Raoult's Law: $\frac{P^o - P_s}{P^o} \approx \frac{n_2}{n_1} = \frac{\Delta T_b \times M_1(\text{kg})}{K_b}$
   • Using $M_1$ in g/mol, the relationship becomes: $M_1(\text{kg}) = M_1(\text{g/mol}) / 1000$. So, $\frac{P^o - P_s}{P^o} = \frac{\Delta T_b \times M_1(\text{g/mol})}{1000 \times K_b}$

5. Calculate the Vapour Pressure of the Solution ($P_s$):

   • Substitute the given values into the derived equation:
     • $P^o = 760$ mm Hg
     • $\Delta T_b = 2$ K
     • $M_1 = 18$ g/mol
     • $K_b = 0.76$ K kg mol⁻¹ $\frac{760 - P_s}{760} = \frac{2 \times 18}{1000 \times 0.76}$ $\frac{760 - P_s}{760} = \frac{36}{760}$
   • From this, we can directly see that: $760 - P_s = 36$
   • Solving for $P_s$: $P_s = 760 - 36$ $P_s = 724 \text{ mm of Hg}$

6. Conclusion: The calculated vapour pressure of the solution is 724 mm of Hg. This matches option A.