Question

At an elevated location with an atmospheric pressure of 0.970 atm, how much table salt (NaCl in grams) do I have to add to 1 L of water in order to be able to boil eggs at 100°C?

Answer 1

According to Clausius Clapeyron equation,

ln(P₂ / P₁) = (H_vap / R) * (1/T₁ - 1/T₂)

where

P₁ = initial pressure = 0.970 atm

P₂ = final pressure = 1.00 atm

T₁ = temperature at P₁

T₂ = temperature at P₂ = 373 K

H_vap = enthalpy of vaporization of water = 40700 J/mol

R = constant = 8.314 J/mol-K

Substituting the values.

ln(1.00 atm / 0.970 atm) = (40700 J/mol / 8.314 J/mol-K) * (1/T₁ - 1/373 K)

(1/T₁ - 1/373 K) = 6.22 x 10^-6 K^-1

1/T₁ = 2.69 x 10^-3 K^-1

T₁ = 372 .14 K

T₁ = 99.14 ^oC

This is the temperature at which pure water will boil at 0.970 atm

Elevation in boiling point = boiling point of solution - boiling point of pure water

Elevation in boiling point = 100 ^oC - 99.14 ^oC

Elevation in boiling point = 0.86 ^oC

molality of NaCl = (Elevation in boiling point) / [(i) * (K_b)]

where

i = van't Hoff factor = 2 (for NaCl)

K_b = 0.512 ^oC/m for water

molality of NaCl = (0.86 ^oC) / [(2) * (0.512 ^oC/m)]

molality of NaCl = 0.843 m

moles NaCl dissolved = (molality of NaCl) * (mass of water in kg)

moles NaCl dissolved = (0.843 m) * (1 kg)

moles NaCl dissolved = 0.843 mol

mass NaCl dissolved = (moles NaCl dissolved) * (molar mass NaCl)

mass NaCl dissolved = (0.843 mol) * (58.44 g/mol)

mass NaCl dissolved = 49.3 g

Hence, 49.3 g of table salt has to be added