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

Jul 15, 2016

The boiling point of the solution is ${102.39}^{o} C$.

#### 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 $C a C {l}_{2}$ dissociates, you obtain three ions.

$C a C {l}_{2} \rightarrow C {a}^{2 +} + 2 C {l}^{-}$

You have one calcium ion and two chloride ions.

Now we can plug in what we know and solve for $\Delta {T}_{b}$:

$\Delta {T}_{b} = \frac{{0.51}^{o} C}{\cancel{m}} \times 1.56 \cancel{m} \times 3$

$\Delta {T}_{b} = {2.39}^{o} C$ This is not your answer. This just tells you how much the boiling point will increase by

Now, we add ${100}^{o} C$ to the $\Delta {T}_{b}$, to determine the new temperature at which the water will boil:

${100}^{o} C + {2.39}^{o} C = {102.39}^{o} C$

***We added ${100}^{o} C$ to that value because ${100}^{o} C$ is the boiling point of water. *