For a dilute solution containing 2.5 gm of a non-volatile non-electrolyte solute in 100 gm of water, the elevation in boiling point at 1 atm pressure is 2oC. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is (Kb=0.76 K kg mol⁻¹)

ΔTb=Kb×m
where ΔTb is the elevation in boiling point, Kb is the ebullioscopic constant, and m is the molality of the solution.
Given:
ΔTb = 2oC
Kb = 0.76 K kg mol⁻¹
Mass of solute = 2.5 g
Mass of solvent (water) = 100 g = 0.1 kg
Assuming the molar mass of the solute is M g/mol, the number of moles of solute is 2.5/M moles.
Molality (m) = moles of solute / kg of solvent = (2.5/M) / 0.1 = 25/M mol/kg
Substituting the values in the equation:
2 = 0.76 × (25/M)
M = (0.76 × 25) / 2 = 9.5 g/mol
Now, let's find the mole fraction of the solute.
Mole fraction of solute (Xsolute) = moles of solute / (moles of solute + moles of water)
Assuming the molar mass of the solute is 9.5 g/mol, number of moles of solute = 2.5 g / 9.5 g/mol ≈ 0.263 moles
Number of moles of water = 100 g / 18 g/mol ≈ 5.56 moles
Xsolute = 0.263 / (0.263 + 5.56) ≈ 0.045
The relationship between vapor pressure lowering and mole fraction is given by Raoult's Law:
ΔP = P°solvent × Xsolute
where ΔP is the vapor pressure lowering, P°solvent is the vapor pressure of pure solvent, and Xsolute is the mole fraction of solute.
At 1 atm pressure and 100°C, the vapor pressure of pure water is approximately 760 mm Hg.
ΔP = 760 mm Hg × 0.045 ≈ 34.2 mm Hg
The vapor pressure of the solution is:
Psolution = P°solvent - ΔP = 760 - 34.2 ≈ 725.8 mm Hg
The closest option is 724 mm Hg.