- Molal boiling point elevation constant of water (Kb) = 0.52 K.kg.mol-1
- 1 atm = 760 mm Hg
- Molar mass of water = 18 g.mol-1
First, we calculate the molality (m) of the solution using the boiling point elevation formula:
ΔTb=Kb×m
From the problem, we know:
2=0.52×m
Solving for m:
m=0.522≈3.846 mol/kg
Next, considering the solution is highly diluted, we use the formula for relative lowering of vapor pressure:
P0ΔP=nsolventnsolute
Given that molality (m) is defined as the number of moles of solute per kilogram of solvent:
P0ΔP=1000m×Msolvent
Substitute the known values:
ΔP=P0×1000m×Msolvent
=760×10003.846×18
=52.615 mm Hg
Therefore, the vapor pressure of the solution:
Psolution=P0−ΔP
=760−52.615
≈707.385 mm Hg
Rounding to the nearest integer, the vapor pressure of the aqueous solution is 707 mm Hg.