Answer:
Vapor pressure of solution = 25.1 Torr
Explanation:
Vapor pressure lowering is one of the colligative property. Its formula is:
ΔP = P° . Xm . i
ΔP = Vapor pressure pure solvent - Vapor pressure solution
P° is Vapor pressure pure solvent
Xm is the mole fraction of solute (mol of solute/ total moles)
i is the Van't Hoff factor, the number of ions particles dissolved.
NaCl → Na⁺ + Cl⁻ (For this dissociation, i = 2)
Let's determine the mole fraction of this solution.
5.50% by mass is 5.50 g of solute in 100 g of solution.
Total mass = 100 g → Mass of solute + Mass of solvent
100 g- 5.5g = 94.5 g (mass of solvent)
Moles of solute → Solute mass / molar mass
5.5 g / 36.45 g/mol = 0.151mol
Moles of solvent → Solvent mass / molar mass
94.5 g / 18 g/mol = 5.25mol
Total moles = Moles of solute + Moles of solvent
0.151 mol + 5.25mol = 5.401 total moles
Xm solute = 0.151 / 5.401 → 0.028
Vapor pressure of water at 25°C → 23.8 Torr
Let's replace the data
Vapor pressure of solution - 23.8 Torr = 23.8 Torr . 0.028 . 2
Vapor pressure of solution = 23.8 Torr . 0.028 . 2 + 23.8 Torr
Vapor pressure of solution = 25.1 Torr
