A common practice in cooking is the addition of salt to boiling water (Kb = 0.52 °C kg/mole). One of the reasons for this might be to raise the temperature of the boiling water. If 2.85 kg of water is boiling at 100 °C, how much NaCl (MW = 58.44 g/mole) would need to be added to the water to increase the boiling point by 2 °C?

281 g of NaCl must be added to 2.5kg of H₂O, to increase the boiing point by 2°C

Explanation:

To solve this question, we apply the concept of boiling point elevation, where the boiling poinf of solution is higher than the boiling point of pure solvent. Formula → ΔT = Kb . m . i

As every colligative property, it depends on the solute

ΔT = Boiling T° of solution - Boiling T° of pure solvent

Kb = Ebulloscopic constant → 0.52 °C/m

m = moles of solute in 1kg of solvent → molality

i = Van't Hoff factor (number of ions dissolved)

Solvent → Water →  Boiling point 100°C

Solute → NaCl. This solute dissociates like this: NaCl → Na⁺ + Cl⁻ (i = 2)

If the water has to increase the boiling point by 2 °C, the ΔT = 2°C. Boiling point of solution (water + NaCl) will be 102°C

Let's replace the data: 2°C = 0.52°C/m . m . 2

1°C / 0.52 m/°C = m → 1.923 mol/kg

1.923 are the moles of NaCl in 1kg, but our mass of solvent is 2.5kg

1.923 mol/kg . 2.5 kg = 4.81 moles (These are the moles of NaCl, we must add to the solution. Let's find out the mass of those moles)

4.81 mol . 58.44 g / 1mol = 281 g

answerambulatory;