Answer: 0.64°C.
Step-by-step explanation:
The boiling point elevation of a solution with a non volatile solute is a colligative property.
A colligative property is a physical property of a solution that depends on the number of dissolved particles of solute.
As such, the elevarion of the boiling point may be estimated by the equation:
ΔTb = i × Kb × m, where:
- ΔTb is the elevation of the boiling point,
- i is the Van't Hoof factor for ionic compounds and is equal to the number of ions produced by one unit formula of compound. For NaCl it is 2: 1 for Na⁺ and 1 for Cl⁻,
- Kb is the molal constant of the solvent, which is given: is 0.52 °C/m
- m is the molality.
To use the formula we must calculate the molality of 3.5% seawater, considereing that the solute is NaCl.
Calculation of molality:
- Formula: m = number of moles of solute / kg of solvent
- basis: 100 g of solution (this is the easiest way to compute m when you know mass %)
- molar mass of NaCl: 58.44 g/mol
- number of moles = mass in grams / molar mass = 3.5g / 58.44g/mol = 0.060 moles
- mass of solvent = 100 g solution - 3.5 g solute = 97.5 g of solvent
- mass of solvent in kg = 97.5g / (1000 g/kg) = 0.0975 kg
- m = 0.060 moles of NaCl / 0.0975 kg of water = 0.62 m
Calculation of ΔTb:
ΔTb = 2 × 0.52°C/m × 0.62 m = 0.64 °C.