First, we need to calculate the molality because that is what we use in our equation for freezing point depression. We can get that from the molarity without knowing exactly how many liters or grams we have. We just have to know what we have one mole per liter. The weight of water is one kilogram per liter, so this allows us to make this conversion.

The molality is 2m. The van't Hoff factor is 3, as we get one calcium ion and two chloride ions per molecule during dissociation.

We can now plug the values into the equation for freezing point depression.

This gives us our depression of .  The normal freezing point of pure water is , which means our new freezing point is .

We can determine how much sodium chloride was added to the water using the freezing point depression equation.

The normal freezing point of wtaer is 0 degrees Celsius, so we know that the temperature of the solution has changed by 2.5 degrees. Since sodium chloride will generate two ions per molecule in solution, the van't Hoff factor will be 2. Based on the density of water, we can determine that 4 liters of water weighs 4 kilograms. 

In order to solve this problem, we can use the boiling point elevation equation: .

We know the temperature change, we can compute molality from the given information, and we know the van't Hoff factor (expected to be 2 in this scenario due to NaCl becoming 2 ions in solution). We can calculate the boiling point constant for the solvent.

We can use the boiling point elevation equation in order to determine the new boiling point once the salt has been added:

Since sodium chloride will form two ions for each molecule in solution, the value for the van't Hoff factor will be 2. In addition, the mass of the water in the solution will be 3 kilograms, which can be determined by using the density of water.

Since water boils at 100 degrees Celsius, this means that the final boiling point of the solution is 100.29 degrees Celsius.

Since the solute is nonvolatile, it does not have an additive vapor pressure, but it will lower the vapor pressure due to taking up surface space in the solution. Using Raoult's law, we can determine the new vapor pressure of the solution:

Here,  is the molar fraction of the water, and  is the original vapor pressure of the water. Since 3 moles of solute were added, water now makes up 80% of the moles in the solution:

Since both compounds are volatile, they will both contribute to the vapor pressure. Using the percentages of each compound in solution, we can determine the vapor pressure for the mixture:

Parts  and  of the equation represent the molar fractions of the compounds multiplied by the corresponding vapor pressures. Adding these together will equal the new vapor pressure of the solution: