Question

What is the boiling point (in °C) of a 1.56 m aqueous solution of `CaCl_2`?

Answer 1

The boiling point of the solution is `102.39^oC`.

Explanation:

For this question we have to use the boiling point elevation equation:

Now, let's determine our known and unknown variables:

Known:
molality
Van't Hoff factor
Molal boiling point constant (we're given this value)

Unknown:
Change in boiling point

I should mention that the Van't Hoff factor basically reflects the number of ions produced in solution upon dissociation of an ionic compound. The compound can also be a molecular one, but the Van't Hoff factor for that is just 1 because molecular compounds do not produce ions in solution.

When `CaCl_2` dissociates, you obtain three ions.

`CaCl_2 rarr Ca^(2+) + 2Cl^(-)`

You have one calcium ion and two chloride ions.

Now we can plug in what we know and solve for `DeltaT_b`:

`DeltaT_b = (0.51^oC)/cancelmxx1.56cancelmxx3`

`DeltaT_b = 2.39^oC` This is not your answer. This just tells you how much the boiling point will increase by

Now, we add `100^oC` to the `DeltaT_b`, to determine the new temperature at which the water will boil:

`100^oC + 2.39^oC = 102.39^oC`

***We added `100^oC` to that value because `100^oC` is the boiling point of water. *