Seawater is about 3.5% by mass dissolved solids, mostly nacl. if `delta"t"_"b" = "ik"_"b" "m"` and kb for water is 0.52°c/m, what is the boiling point elevation of seawater?

0.64°C.

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/mm 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 solventbasis: 100 g of solution (this is the easiest way to compute m when you know mass %)mass of solute: 3.5 gmolar mass of NaCl: 58.44 g/molnumber of moles = mass in grams / molar mass = 3.5g / 58.44g/mol = 0.060 molesmass of solvent = 100 g solution - 3.5 g solute = 97.5 g of solventmass of solvent in kg = 97.5g / (1000 g/kg) = 0.0975 kgm = 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.

50 meters per 5 seconds

The answer is b hope this