30. 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

delta T = i*Kb*m

i = 2 for NaCl
i = 3 for CaCl2
Kb and m are the same for both NaCl and CaCl2.

To compare the boiling points of the two solutions, we need to understand the concept of boiling point elevation, which is a colligative property.

The boiling point of a liquid is the temperature at which its vapor pressure becomes equal to the atmospheric pressure. When a solute is added to a solvent, it affects the vapor pressure and subsequently raises the boiling point.

According to Raoult's law, the vapor pressure of a solvent above a solution is directly proportional to the mole fraction of the solvent present. As the mole fraction of the solvent decreases due to the addition of a solute, the vapor pressure decreases as well. Consequently, the boiling point of the solution increases compared to the pure solvent.

Now, let's apply this knowledge to the given scenario. In one liter of water, we have 1 mole of NaCl, which dissociates into one sodium ion (Na+) and one chloride ion (Cl-). In the case of CaCl2, we still have one liter of water, but now we have 1 mole of CaCl2, which dissociates into one calcium ion (Ca2+) and two chloride ions (2Cl-).

The number of moles of particles in the solution affects the extent of boiling point elevation. Since CaCl2 produces more ions (three in total) compared to NaCl (two in total), the boiling point of water with CaCl2 will be higher.

Therefore, the boiling point of a liter of water containing 1 mole of calcium chloride (CaCl2) will be higher than that of a liter of water containing 1 mole of sodium chloride (NaCl).