At 2°C, the vapor pressure of pure water is 23.76 mmHg and that of a certain seawater sample is 23.09 mmHg. Assuming that seawater contains only NaCl, estimate its molal concentration.

Molal concentration of NaCl in Seawater = 1.6 moles of NaCl/kg of pure water!!!

Explanation:

Let the mole fraction of pure water be X, that of NaCl = 1 - X

We obtain the mole fraction of solvent from Raoult's law relation

Pressure of solution, P⁰ₛₐₗₜ = Vapour pressure of solvent, P⁰ₕ₂₀ × mole fraction of solvent, X

23.09 = 23.76 × X

X = 0.972

Using a mole basis of 1 mole,

Number of moles of pure water = 0.972

number of moles of NaCl in the seawater sampler = 1 - 0.972 = 0.028 moles

Molal concentration = number of moles of solute/mass of solvent in kg

Mass of solvent = number of moles × molar mass of pure water = 0.972 × 18 = 17.496g = 0.0175 kg

Molal concentration = 0.028/0.0175 = 1.6 moles of NaCl/kg of pure water!!!

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