How will the boiling point of a liter of water containing 1 mole of sodium chloride (NaCl) compare with that of a liter of water containing 1 mole of calcium chloride (CaCl2)? Explain your answer.
Look at the formula for boiling point of each.
delta T = i*Kb*molality
Molality is the same for both solution.
Kb is the same for both solution.
There delta T depends upon i, the van't Hoff factor.
i = 2 for NaCl
i = 3 for CaCl2.
To compare the boiling points of two solutions, we need to consider the concept of boiling point elevation. Boiling point elevation occurs when a solute (in this case, either sodium chloride or calcium chloride) is dissolved in a solvent (water), resulting in the boiling point of the solution being higher than that of the pure solvent.
The boiling point elevation is directly proportional to the molality of the solute. Molality (m) is the ratio of the number of moles of solute (n) to the mass of the solvent (in kg), given by the equation:
m = n / (molar mass of solvent * mass of solvent in kg)
In this case, both solutions contain 1 mole of either sodium chloride (NaCl) or calcium chloride (CaCl2) and a liter of water, which has a mass of approximately 1 kg. However, since calcium chloride (CaCl2) dissociates into three ions in water (1 Ca2+ ion and 2 Cl- ions), while sodium chloride (NaCl) dissociates into two ions (1 Na+ ion and 1 Cl- ion), the molality of calcium chloride will be higher than that of sodium chloride.
Due to the greater number of particles (ions) in the calcium chloride solution, it will exhibit a higher boiling point elevation, and therefore its boiling point will be higher than that of the sodium chloride solution.
Therefore, the boiling point of a liter of water containing 1 mole of sodium chloride (NaCl) will be lower than that of a liter of water containing 1 mole of calcium chloride (CaCl2).