Assuming that sea water is a 3.5 mass % solution of nacl and that its density is 1.00 g/ml, calculate its boiling point in ∘c.

Answer is: 100,635°C.
ω(NaCl) = 3,5% = 0,035.
d(sea water) = 1 g/ml.
Tb(water) = 100°C.
Tb(sea water) = ?
Kb(water) = 0.512 °K·kg/mol, Kb - ebullioscopic constant.
m(Nacl) = 0,035·1000g = 35g.
m(water) = 1000g - 35g = 965g = 0,965kg.
n(NaCl) = 35g ÷ 58,5 g/mol = 0,6 mol.
b(NaCl) = n(NaCl) ÷ m(H₂O) = 0,62 mol/kg.
ΔT = Kb · b · i.
i - van 't Hoff factor, for NaCl (i=2).
ΔT = 0,635°C.
T(sea water) = T(water) + ΔT = 100,635°C.

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